LECTURES ON MATHEMATICAL ANALYSIS FOR ECONOMISTS · 2021. 3. 23. · LECTURES ON MATHEMATICAL...

LECTURES ON MATHEMATICAL ANALYSISFOR ECONOMISTS

Tapan MitraCornell University

WORKED OUT SOLUTIONS TO PROBLEM SETSChristopher HandyCornell University

August 2011Economics 6170

Contents

I Linear Algebra 1

1 Vectors 21.1 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21.2 Linear Dependence of Vectors . . . . . . . . . . . . . . . . . . . . . . .. . . 31.3 Rank and Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Inner Product and Norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 51.5 Worked Out Problems on Chapter 1 . . . . . . . . . . . . . . . . . . . . . .. 7

2 Matrices 152.1 Matrix Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15

2.1.1 Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.1.2 Some Words of Caution . . . . . . . . . . . . . . . . . . . . . . . . . 162.1.3 Transpose of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 172.1.4 Some Special Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Rank of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 182.4 Relationship Between Invertible and Non-Singular Matrices . . . . . . . . . 182.5 Worked Out Problems on Chapter 2 . . . . . . . . . . . . . . . . . . . . . .. 21

3 Simultaneous Linear Equations 283.1 System of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . .. . . 283.2 Existence of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 283.3 Uniqueness of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 313.4 Calculation of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 323.5 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .323.6 Matrix Inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 343.7 Cramer’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.8 Appendix: System of Homogeneous Linear Equations . . . . .. . . . . . . 37

1

CONTENTS 2

3.9 Worked Out Problems on Chapter 3 . . . . . . . . . . . . . . . . . . . . . .. 40

4 Characteristic Values and Vectors 474.1 The Characteristic Value Problem . . . . . . . . . . . . . . . . . . . .. . . . 474.2 Characteristic Values, Trace and Determinant of a Matrix. . . . . . . . . . . 484.3 Characteristic Values and Vectors of Symmetric Matrices. . . . . . . . . . . 504.4 Spectral Decomposition of Symmetric Matrices . . . . . . . .. . . . . . . . 514.5 Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .534.6 Characterization of Quadratic Forms . . . . . . . . . . . . . . . . .. . . . . . 544.7 Alternative Characterization of Quadratic Forms . . . . . .. . . . . . . . . . 554.8 Appendix: Spectral Decomposition of Non-Symmetric Matrices . . . . . . 574.9 Worked Out Problems on Chapter 4 . . . . . . . . . . . . . . . . . . . . . .. 59

II Real Analysis 69

5 Basic Concepts of Real Analysis 705.1 Norm and Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .705.2 Open and Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 715.3 Convergent Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 725.4 Compact Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 735.5 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 735.6 Existence of Solutions to Constrained Optimization Problems . . . . . . . . 745.7 Appendix I: Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 755.8 Appendix II: Continuity of Functions of Several Variables . . . . . . . . . . 775.9 Appendix III: On a Variation of Weierstrass Theorem . . . .. . . . . . . . . 795.10 Worked Out Problems on Chapter 5 . . . . . . . . . . . . . . . . . . . . .. . 81

6 Differential Calculus 876.1 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 876.2 Composite Functions and the Chain Rule . . . . . . . . . . . . . . . . . .. . 886.3 Homogeneous Functions and Euler’s Theorem . . . . . . . . . . .. . . . . . 896.4 The Inverse and Implicit Function Theorems . . . . . . . . . . .. . . . . . . 926.5 Worked Out Problems on Chapter 6 . . . . . . . . . . . . . . . . . . . . . .. 97

7 Convex Analysis 1077.1 Convex Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1077.2 Separating Hyperplane Theorem for Convex Sets . . . . . . . . .. . . . . . 1087.3 Continuous and Differentiable Functions on Convex Sets . .. . . . . . . . . 110

CONTENTS 3

7.4 Concave Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1117.5 Quasi-Concave Functions . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 1137.6 Worked Out Problems on Chapter 7 . . . . . . . . . . . . . . . . . . . . . .. 116

III Classical Optimization Theory 126

8 Unconstrained Optimization 1278.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1278.2 Necessary Conditions for a Local Maximum . . . . . . . . . . . . . .. . . . 1288.3 Sufficient Conditions for a Local Maximum . . . . . . . . . . . . . .. . . . 1298.4 Sufficient Conditions for a Global Maximum . . . . . . . . . . . . .. . . . . 1308.5 The Method of Least Squares . . . . . . . . . . . . . . . . . . . . . . . . .. . 1318.6 The Envelope Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1338.7 Worked Out Problems on Chapter 8 . . . . . . . . . . . . . . . . . . . . . .. 136

9 Constrained Optimization 1419.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1419.2 Necessary Conditions for a Constrained Local Maximum . . . .. . . . . . . 1419.3 The Arithmetic Mean-Geometric Mean Inequality . . . . . . .. . . . . . . . 1439.4 Sufficient Conditions for a Constrained Local Maximum . . . .. . . . . . . 1449.5 Sufficient Conditions for a Global Maximum . . . . . . . . . . . . .. . . . . 1459.6 Worked Out Problems on Chapter 9 . . . . . . . . . . . . . . . . . . . . . .. 147

IV Modern Optimization Theory 153

10 Concave Programming 15410.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 15410.2 Constrained Global Maxima and Saddle Points . . . . . . . . . .. . . . . . . 15510.3 The Kuhn-Tucker Conditions and Saddle Points . . . . . . . . .. . . . . . . 15810.4 The Kuhn-Tucker Conditions and Constrained Local Maxima. . . . . . . . 15910.5 Constrained Local and Global Maxima . . . . . . . . . . . . . . . . .. . . . 16310.6 Appendix: On the Kuhn-Tucker Conditions . . . . . . . . . . . . .. . . . . . 16410.7 Worked Out Problems on Chapter 10 . . . . . . . . . . . . . . . . . . . .. . 166

11 Quasi-Concave Programming 18111.1 Properties of Concave and Quasi-concave functions . . . .. . . . . . . . . . 181

11.1.1 Concave Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

CONTENTS 4

11.1.2 Quasi-concave functions . . . . . . . . . . . . . . . . . . . . . . .. . 18211.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .18311.3 The Sufficiency Theorem of Arrow-Enthoven . . . . . . . . . . .. . . . . . 18411.4 The Necessity Theorem of Arrow-Enthoven . . . . . . . . . . . .. . . . . . 18611.5 Worked Out Problems on Chapter 11 . . . . . . . . . . . . . . . . . . . .. . 188

12 Linear Programming 19312.1 The Primal and Dual Problems . . . . . . . . . . . . . . . . . . . . . . .. . . 19312.2 Optimality Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 19412.3 The Basic Duality Theorems . . . . . . . . . . . . . . . . . . . . . . . . .. . 19412.4 Complementary Slackness . . . . . . . . . . . . . . . . . . . . . . . . . .. . 19712.5 Worked Out Problems on Chapter 12 . . . . . . . . . . . . . . . . . . . .. . 199

Part I

Linear Algebra

1

Chapter 1

Vectors

1.1 Vector Spaces

In defining vector spaces, we will consider thefield to be given by the set of reals, denotedbyR. [One can define this set formally, but we will not do so here.]The elements ofR arecalledscalarsor numbers.

An m-vector xis an ordered set ofm numbers(x1, ...,xm). The numberxi is calledthe ith coordinateof x. We use the notationx = (xi), meaningx is the vector whoseithcoordinate isxi.

The set of allm-vectors is calledm-spaceand is denoted byRm.Some Special Vectors:

Theith unit vectoris the vector whose ith coordinate is 1, and whose other coordinatesare zero. We denote the ith unit vector byei. Thesum vector, denoted byu, is the vectorall of whose coordinates are 1. Thenull vector, denoted by 0, is the vector all of whosecoordinates are 0.Vector Operations:

Two m-vectors,x andy, are said to beequal(written x= y) if xi = yi for i = 1, ...,m.We now define two algebraic operations on the vectors ofRm.

Addition:If x= (xi) andy= (yi) arem-vectors, theirsum x+y is the vector (xi +yi).

Scalar Multiplication:If x= (xi) is anm-vector andλ is a number, theproductλx is the vector (λxi).Given these two definitions, a number of properties, listed below, follow immediately.

For addition, we have:(A.1) (x+y)+z= x+(y+z) [Associative Law](A.2) x+y= y+x [Commutative Law]

2

CHAPTER 1. VECTORS 3

(A.3) For everyx andy, there iszsuch thatx + z= y [Law of Subtraction]

For multiplication, we have:(M.1) λ(x+y) = λx+λy [Vector Distributive Law](M.2) (λ+µ)x= λx+µx [Scalar Distributive Law](M.3) λ(µx) = (λµ)x [Scalar Associative Law](M.4) 1x= x [Identity Law]

The properties listed above may be taken as axioms for an abstract algebraic system.Such systems are calledvector spaces.

The vector space that we will study consists of a field,R; them-space,Rm; the oper-ations of addition and scalar multiplication. We will generally refer to this vector spaceitself simply asRm, although this is clearly a shorthand.

1.2 Linear Dependence of Vectors

A set of vectorsx1, ...,xn in Rm is linearly dependentif there exist numbersλ1, ...,λn, not

all zero, such that

n∑i=1

λixi = 0

If the vectors are not linearly dependent, they are calledlinearly independent.A vectory is a linear combinationof the vectorsx1

, ...,xn if

y= n∑i=1

λixi

for some numbersλi ∈ R. A set of vectorsx1, ...,xn spansRm if every y ∈ Rm can be

expressed as a linear combination ofx1, ...,xn.

Theorem 1. (Fundamental Theorem on Vector Spaces): If each of the vectors y0, y1

, ...,ym

in the vector spaceRn is a linear combination of the vectors x1, ...,xm, then the yi are

linearly dependent.

Corollary 1. Any set of(m+1) vectors inRm is linearly dependent.


Proof. For any vectorx in Rm,

x= m∑i=1

xiei

Consequently, for any set of(m+1) vectors inRm, we can write each as a linear combi-nation of the vectors(e1

, ...,em). So the set of(m+1) vectors must be linearly dependentby Theorem 1.

Corollary 2. Any system of m homogeneous linear equations in(m+1) unknowns has anon-zero solution.

Proof. We can write the system ofm homogeneous linear equations in(m+1) unknownsin the following way:

a10x0+a11x1+a12x2+ ....+a1mxm= 0.................................................................

am0x0+am1x1+am2x2+ ....+ammxm= 0

⎫⎪⎪⎪⎬⎪⎪⎪⎭(1.1)

Definea j = (a1 j , a2 j , ....,am j) for j = 0, ...,m. Then thea j are a set of(m+1) vectorsin m-space; hence by Corollary 1, they are linearly dependent. Sothere exist numbersλ0, ....,λm, not all zero, such that

m∑j=0

λ ja j = 0

Then(x j) = (λ j) gives the desired solution of (1.1).

1.3 Rank and Basis

Let Sbe a subset of the vector spaceRm. Therankof Sis the maximum number of linearlyindependent vectors which can be chosen fromS.

If r is the rank ofS, a set ofr linearly independent vectors ofS is called abasisfor S.

Corollary 3. Rm has rank m.

Proof. By Corollary 1, the rank ofRm is at mostm. On the other hand, the unit vectorse1, ...,em are clearly linearly independent, so the rank ofRm is at leastm.

Theorem 2. (Basis Theorem) Suppose x1, ...,xr are linearly independent vectors in S.

Then x1, ...,xr is a basis for S if and only if every vector y in S is a linear combinationof the vectors x1, ...,xr

.


Proof. Suppose everyy ∈S is a linear combination of thexi. Then any set of more thanrvectors inS is linearly dependent by Theorem 1, and soShas rankr, and thexi are a basis.

Suppose thexi form a basis ofS. Then,Shas rankr; so if y ∈S, the vectorsx1, ...,xr

,yare linearly dependent. Thus there existλ0,λ1, ...,λr in R, not all zero, such that

r∑i=1

λixi +λ0y= 0

Clearly λ0 ≠ 0, otherwise thexi are linearly dependent. Hence, definingµi = −λi/λ0 fori = 1, ...,r, we have

y= r∑i=1

µixi

soy is a linear combination ofx1, ...,xr .

1.4 Inner Product and Norm

We now introduce a third operation inRm. If x andy are vectors inRm, their inner productis denoted byxyand is defined as the number:

xy= m∑i=1

xiyi

Let x be anm-vector. The inner product ofx and the ith unit vectorei is

xei = xi f or i = 1, ...,m

The inner product ofx and the sum vectoru is

xu= m∑i=1

xi

Some properties of the inner product are listed below:

(I.1) xy= yx (Commutative Law)(I.2) (λx)y= λ(xy) (Mixed Associative Law)(I.3) (x+y)z= xz+yz (Distributive Law)(I.4) x2 ≡ xx is 0 if and only ifx= 0

If x∈Rm, then the (Euclidean)normof x, denoted by∥x∥ , is the (non-negative) number


∥x∥ = [ m∑i=1

x2i ]

1/2

Thus the norm is the (non-negative) square root of the inner product ofx with itself.The following properties of the norm can be verified:

(N.1) ∥x∥ = 0 if and only ifx= 0(N.2) ∥λx∥ = ∣λ∣ ∥x∥(N.3) ∥x+y∥ ≤ ∥x∥+∥y∥Property (N.3) is usually referred to as the “triangle inequality” for norm.

Two vectorsx andy are calledorthogonalif their inner product is zero; that is, ifxy=0.For orthogonal vectors, we have the Pythagoras theorem:

(N.4) If xy= 0, then∥x+y∥2 = ∥x∥2+∥y∥2If x andy are orthogonalm-vectors, they are calledorthonormalif∥x∥ = ∥y∥ = 1.


1.5 Worked Out Problems on Chapter 1

Problem 1 (Linear Dependence and Independence).

Let S= {e1,e2

, . . . ,en} be the set of unit vectors inRn.(a) LetT = {x1

,x2, . . . ,xn} be a set of vectors inRn

, defined by:

x1 = e1+e2,x2 = e2+e3

,⋯,xn−1 = en−1+en,xn = en+e1

.

Is T a set of linearly independent vectors inRn? Explain. [Hint: consider two cases,nodd,n even].

(b) Let x be an arbitrary vector inRn, with xn ≠ 0. LetU be the set of vectors defined

by U = {e1,e2

,⋯,en−1,x}. IsU a set of linearly independent vectors inRn? Explain.

Solution.

(a) We will show that the vectorsx1, . . . ,xn are linearly dependent whenn is even and

linearly independent whenn is odd.

Consider a linear combination of the vectors ofT:

λ1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1100⋮000

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

+λ2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0110⋮000

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

+ ⋅ ⋅ ⋅+λn−1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0000⋮011

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

+λn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1000⋮001

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

λn+λ1

λ1+λ2

λ2+λ3

λ3+λ4⋮λn−3+λn−2

λn−2+λn−1

λn−1+λn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦If n is even, we can findλ1, . . . ,λn not all zero such that this linear combination givesthe zero vector:

λi =⎧⎪⎪⎨⎪⎪⎩

1, if i odd

−1, if i evenfor all i = 1, . . . ,n

Thus forn even, the vectors ofT are linearly dependent.

But if n is odd, theseλi do not give the zero vector, because thenλn+λ1 ≠ 0. In fact,we can show that there are noλ1, . . . ,λn not all zero such that the linear combinationabove is equal to the zero vector. Suppose there areλ1, . . . ,λn not all zero such that

⎡⎢⎢⎢⎢⎢⎢⎢⎣

λn+λ1

λ1+λ2⋮λn−1+λn

⎤⎥⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎢⎣

00⋮0

⎤⎥⎥⎥⎥⎥⎥⎥⎦


Then, reading from the last equation to the first, sincen is odd it must be that

λn = −λn−1 = λn−2 = ⋅ ⋅ ⋅ = −λ2 = λ1.

Since theλ1, . . . ,λn are not all zero, it must be that eachλi ≠ 0, so thatλn+λ1 = 2λ1 ≠0. This is a contradiction, so we conclude that whenn is odd, the vectors ofT arelinearly independent.

(b) We can prove the vectors ofU are linearly independent by contradiction. Supposethey are linearly dependent, so that there areµ1, . . . ,µn not all zero such that

µ1e1+⋅ ⋅ ⋅+µn−1en−1+µnx=⎡⎢⎢⎢⎢⎢⎢⎢⎣

µ1+µnx1⋮µn−1+µnxn−1

µnxn

⎤⎥⎥⎥⎥⎥⎥⎥⎦= 0.

Sincexn≠0, from the last row we must haveµn=0. But then the equation above onlyholds if µi = 0 for all i = 1, . . . ,n. This contradicts the assumption that theµ1, . . . ,µn

are not all zero, so we conclude that the vectors ofU are linearly independent.

Problem 2 (Rank).

(a) LetSbe a set of vectors inRn, defined by:

S= {(x1, ...,xn) ∈Rn ∶ x1+x2+⋯+xn = 2}What is the rank ofS? Explain.

(b) LetT be a set of vectors inRn, defined by:

T = {(x1, ...,xn) ∈Rn ∶ x1+2x2+⋯+nxn = 0}What is the rank ofT? Explain.

Solution.

(a) First, note that sinceS⊆ Rn, it must be that rank(S) ≤ n. If we can findn linearlyindependent vectors inS, we will know that rank(S) ≥ n and this will allow us toconclude that rank(S) = n.


Define vectorsdi = 2ei for i = 1, . . . ,n, where theei are unit vectors inRn. Note thateachdi ∈ S. It is easy to check that these vectors are linearly independent: assumethey are not, so that there are someλ1, . . . ,λn not all zero such that

λ1d1+⋅ ⋅ ⋅+λndn =⎡⎢⎢⎢⎢⎢⎣2λ1⋮2λn

⎤⎥⎥⎥⎥⎥⎦= 0.

This implies that eachλi = 0, which is a contradiction. Thus then vectorsd1, . . . ,dn

are linearly independent, and we conclude that rank(S) = n as outlined above.

(b) Let’s try to gain some intuition from the two-dimensional case: consider the setT2 = {(x1,x2) ∈ R2 ∶ x1+2x2 = 0}. Now, rank(T2) = 1 because every vector inT2

can be expressed asx1(1,−12). This suggests that in then-dimensional case, the

restrictionx1+2x2+ ⋅ ⋅ ⋅ +nxn = 0 reduces the rank ofT to n−1. [Note this does nothappen in part (a) because the restriction on the setSdoes not involve a zero on theright hand side. Think about this distinction geometrically in R2.]

To prove thatT, which is a subset ofRn, has rankn−1, we really need to show twothings. First, we must constructn−1 linearly independent vectors fromT. Second,we must show that anyn vectors inT are necessarily linearly dependent. By thedefinition of rank(T) as the maximum number of linearly independent vectors thatcan be chosen fromT, this will show that rank(T) = n−1.

We first claim that then−1 vectors⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

−2100⋮00

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

−3010⋮00

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

, . . . ,

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

−n000⋮01

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦are elements ofT and are linearly independent. These can be written more com-pactly asxi =ei− ie1 for i =2, . . . ,n. It is clear that eachxi satisfiesxi

1+2xi2+⋅ ⋅ ⋅+nxi

n =0. Now, suppose thexi are linearly dependent, so that there areµ2, . . . ,µn not all zerosuch that

µ2x2+⋅ ⋅ ⋅+µnxn =⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

−2µ2−⋅ ⋅ ⋅−nµn

µ2

µ3⋮µn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦= 0.


This can hold only if allµ2, . . . ,µn are equal to zero, which is a contradiction. There-fore thexi, as defined above, constituten−1 linearly independent vectors fromT, sorank(T) ≥ n−1.

Now, it remains only to show that anyn vectors fromT are linearly dependent.So, considern arbitrary vectorsy1

, . . . ,yn, each an element ofT. These are linearlydependent if there existθ1, . . . ,θn not all zero such that

θ1

⎡⎢⎢⎢⎢⎢⎢⎢⎣

−(2y12+⋅ ⋅ ⋅+ny1

n)y1

2⋮y1

n

⎤⎥⎥⎥⎥⎥⎥⎥⎦+ ⋅ ⋅ ⋅+θn

⎡⎢⎢⎢⎢⎢⎢⎢⎣

−(2yn2+⋅ ⋅ ⋅+nyn

n)yn

2⋮yn

n

⎤⎥⎥⎥⎥⎥⎥⎥⎦= 0.

The lastn−1 rows form a system ofn−1 homogeneous linear equations in then unknownsθ1, . . . ,θn, and Corollary 2 tells us this system must have a non-zerosolution in the thetas. Now, let’s look at the first equation in the above system:regroup terms for

−2(θ1y12+⋅ ⋅ ⋅+θnyn

2)− ⋅ ⋅ ⋅−n(θ1y1n+⋅ ⋅ ⋅+θnyn

n) = 0.

For theθ1, . . . ,θn that solve the lastn−1 equations, this equation also holds, becausethe left hand side is zero term by term. Remembering that Corollary 2 ensures theθ1, . . . ,θn are not all zero, we conclude that anyn vectors inT are linearly dependent.This last step shows that rank(T) < n, and together with our conclusion above thatrank(T) ≥ n−1, we finally have the result: rank(T) = n−1.

Note that it is also possible to prove that any vectorz∈ T can be written as a linearcombination of the vectorsx2

, . . . ,xn. By the Basis Theorem, the vectorsx2, . . . ,xn

are then a basis forT, which is only possible if rank(T) = n−1. But the proof aboveis more fundamental in that it avoids the idea of basis altogether.

Problem 3 (Basis).

Let S= {x1, . . . ,xm} be a set of linearly independent vectors inRn

, with m< n, and letT = {y1

, ...,yn} be a set of basis vectors ofRn.

Show that there are(n−m) vectors in the setT, such that them vectors inS, togetherwith these(n−m) vectors inT, constitute a basis ofRn

.

Solution.


The easiest way to prove this is by induction. The idea is thatwe can first find one ofthey j to replace byx1 and the resulting set ofn vectors will still be a basis forRn. We canthen do the same forx2, thenx3, and so on until we have replacedm of they j by all thex1, . . . ,xm and then vectors still form a basis forRn. This works because thexi are linear

combinations of they j , so whatever property of they j is making them a basis is somehowembodied in thexi. We just have to be careful about whichy j we replace with one of thexi.

The base case in the induction proof is to show that there is somey j that can be replaced byx1 so that the remaining vectors inT, together withx1, constitute a basis forRn. Now, notethat none of thexi may be the zero vector; otherwise,Swould not be linearly independent.Similarly, none of they j may be the zero vector. That means that when we use the factthaty1

, . . . ,yn is a basis forRn to write

x1 = λ1y1+⋅ ⋅ ⋅+λnyn, (1.2)

we know thatλ1, . . . ,λn are not all zero. Strictly speaking we do not know whichλi arenonzero, but for notational simplicity we can assume without loss of generality thatλ1 ≠ 0.(Alternatively, you can think of this as reordering they j to fit our notation.) This allowsus to use (1) to write

y1 =1λ1

x1− λ2

λ1y2−⋅ ⋅ ⋅− λn

λ1yn. (1.3)

We want to conclude thatx1,y2

, . . . ,yn are a basis forRn. Since rank(Rn) = n, we onlyneed to show that then vectorsx1

,y2, . . . ,yn are linearly independent. Suppose they are

not, so that there areµ1, . . . ,µn not all zero such that

µ1x1+µ2y2+⋅ ⋅ ⋅+µnyn = 0. (1.4)

Now, µ1 ≠ 0 because ifµ1 = 0 then the vectorsy2, . . . ,yn are linearly dependent, which

contradictsy1, . . . ,yn being a basis forRn. Then we can use (3) to write

x1 = −µ2

µ1y2−⋅ ⋅ ⋅− µn

µ1yn. (1.5)

Substituting (4) into (2), we have thaty1, . . . ,yn are linearly dependent, which contradicts

y1, . . . ,yn being a basis forRn. At last we conclude thatx1

,y2, . . . ,yn are indeed linearly

independent, so they form a basis forRn.

The inductive case in the proof is, fortunately, very similar to the base case. We wantto show that ifx1

, . . . ,xk−1,yk

,yk+1, . . . ,yn is a basis forRn, wherek− 1 < m < n, then

x1, . . . ,xk−1

,xk,yk+1

, . . . ,yn is also a basis forRn, again subject to reordering theyk, . . . ,yn


to suit our notation. Now, since by hypothesisx1, . . . ,xk−1

,yk,yk+1

, . . . ,yn form a basis forRn, and sincexk ≠ 0, there areθ1, . . . ,θn not all zero such that

xk = θ1x1+⋅ ⋅ ⋅+θk−1xk−1+θkyk+θk+1yk+1+⋅ ⋅ ⋅+θnyn

. (1.6)

It is not possible that allθk, . . . ,θn are zero, because thenxk would be a linear combinationof x1

, . . . ,xk−1, which would be a contradiction of the linear independence of S. So, assumewithout loss of generality thatθk ≠ 0. Then we can use (5) to write

yk = −θ1

θkx1−⋅ ⋅ ⋅− θk−1

θkxk−1+ 1

θkxk− θk+1

θkyk+1−⋅ ⋅ ⋅− θn

θkyn. (1.7)

We want to conclude thatx1, . . . ,xk−1

,xk,yk+1

, . . . ,yn are a basis forRn. Since rank(Rn)=n,we only need to show that then vectorsx1

, . . . ,xk−1,xk

,yk+1, . . . ,yn are linearly indepen-

dent. Suppose they are not, so that there areα1, . . . ,αn not all zero such that

α1x1+⋅ ⋅ ⋅+αk−1xk−1+αkxk+αk+1yk+1+⋅ ⋅ ⋅+αnyn = 0. (1.8)

Now, αk ≠ 0 because ifαk = 0 then the vectorsx1, . . . ,xk−1

,yk+1, . . . ,yn are linearly depen-

dent, which contradictsx1, . . . ,xk−1

,yk,yk+1

, . . . ,yn being a basis forRn. Then we can use(7) to write

xk = −α1

αkx1−⋅ ⋅ ⋅− αk−1

αkxk−1− αk+1

αkyk+1−⋅ ⋅ ⋅− αn

αkyn. (1.9)

Substituting (8) into (6), we have thatx1, . . . ,xk−1

,yk,yk+1

, . . . ,yn are linearly dependent,which contradictsx1

, . . . ,xk−1,yk

,yk+1, . . . ,yn being a basis forRn. At last we conclude that

x1, . . . ,xk−1

,xk,yk+1

, . . . ,yn are indeed linearly independent, so they form a basis forRn.

By induction, then, we have that then vectorsx1, . . . ,xm

,ym+1, . . . ,yn are a basis forRn,

after suitable reordering of the originaly j .

Problem 4 (Inner Product).

SupposeS= {x1, . . . ,xn} is a set of non-null vectors inRn, which are mutually orthog-

onal: that is,xix j = 0 wheneveri ≠ j. Show thatS is linearly independent.

Solution.

We will prove by contradiction thatSis linearly independent. Suppose thatSis linearlydependent, so that for someλ1, . . . ,λn not all zero, we haveλ1x1+ ⋅ ⋅ ⋅ +λnxn = 0. Now,


consider somek ∈ {1, . . . ,n} for which λk ≠ 0. We have

0= xk ⋅0= xk ⋅ (λ1x1+⋅ ⋅ ⋅+λnxn) = n∑i=1

λi(xk ⋅xi)= λk(xk ⋅xk) (sincexk ⋅xi = 0 for i ≠ k)

≠ 0 (sinceλk ≠ 0 andxk ≠ 0)

This contradiction establishes thatS is linearly independent.


ReferencesMuch of this material is standard in texts on linear algebra,like Linear Algebraby G.

Hadley(Chapter 2), orElementary Matrix Algebraby Franz Hohn(Chapters 4,5). A goodexposition can also be found inThe Theory of Linear Economic Modelsby David Gale(Chapter 2). You will find a proof of the fundamental theorem onvector spaces (by usingmathematical induction) in Gale’s book.∎

Chapter 2

Matrices

2.1 Matrix Algebra

An m×n matrixis a rectangular array of numbersai j , i =1, ...,m; j =1, ...,n. Thus, we write

A= (ai j ) =⎡⎢⎢⎢⎢⎢⎣

a11..........a1n

.............

am1.........amn

⎤⎥⎥⎥⎥⎥⎦Then-vectorAi = (ai1, ...,ain) is called theith row vectorof A; them-vectorA j = (a1 j , ...,am j)is called thejth column vectorof A. The matrixA hasm row vectorsA1, ...,Am and it hasn column vectorsA1

, ...,An. Thus anm×n matrix may be interpreted as an ordered set ofm row vectors, or as an ordered set ofn column vectors.

In view of this above statement, the operations on matrices follow from the operationson vectors.

2.1.1 Matrix Operations

Two m×n matricesA andB areequal(writtenA=B) if ai j = bi j for i = 1, ...,m; j = 1, ...,n.If A andB arem×n matrices, theirsum A+B is anm×n matrix,(ai j +bi j ).If A is anm×n matrix, andλ is a scalar, theirproductλA is anm×n matrix (λai j ).Let A be anm×n matrix, andB ann× r matrix. ThenB can be premultiplied byA or A

can be post-multiplied byB. Thematrix product, denoted byAB is anm× r matrix givenby

( n∑k=1

aikbk j) i = 1, ...,m; j = 1, ...,r

15

CHAPTER 2. MATRICES 16

For both operationsABandBA to be defined, ifA is m×n, thenB must ben×m.The following properties related to matrix addition and multiplication can be verified

[assuming the relevant matrices can be added and/or multiplied].(MA.1) A+B=B+A [Commutative Law](MA.2) (A+B)+C=A+(B+C) [Associative Law](MM.1) (AB)C=A(BC) =ABC [Associative Law](MM.2) A(B+C) =AB+AC [Distributive Law](B+C)A=BA+CA [Distributive Law]

2.1.2 Some Words of Caution

Some results which are true for real numbers are not necessarily true for matrices. Itis useful to be aware of some of these.

(i) AB is not necessarily equal to BA.

Example 1. Let

A= [1 23 4] , B= [0 −1

6 7]

Then

AB= [12 1324 25

] , BA= [−3 −427 40

](ii) AB= 0 is possible with neitherA norB being the null matrix.

Example 2. Let

A= [2 41 2] , B= [−2 4

1 −2]Then,

AB= [0 00 0]

(iii) CD=CE with C not null is possible withoutD andE being equal.

Example 3. Let

C= [2 36 9] , D = [1 1

1 2] , E = [−2 1

3 2]

Then

CD=CE= [ 5 815 24

] , C≠ 0 and D≠E.


2.1.3 Transpose of a Matrix

If A is anm×n matrix, then then×mmatrix B defined by

bi j = a ji i = 1, ...,n; j = 1, ...,m

is called thetransposeof A, and is denoted byA′.The following properties of transposes can be easily verified:(T.1) (A′)′ =A(T.2) (A+B)′ =A′+B′

(T.3) (AB)′ =B′A′

2.1.4 Some Special Matrices

There are some special types of matrices, which are now discussed below. Anm×n matrixis asquare matrixif m= n. An n×n matrix issymmetricif

ai j = a ji i ≠ j

[That is, a square matrix,A, is symmetric ifA=A′.] An n×n matrix is adiagonal matrixif

ai j = 0 i ≠ j

An n×n matrix is anidentity matrix(denoted byIn or I) if

aii = 1 i = 1, ...,n

ai j = 0 i ≠ j

An m×n matrix is anull matrix (denoted by 0) if

ai j = 0 i = 1, ...,m; j = 1, ...,n.

Note that a null matrix need not be a square matrix.The identity and null matrices are especially useful in matrix algebra. The following

properties of identity and null matrices can be verified (assuming the relevant matrices canbe added and/or multiplied)

(I.1) AI = IA =A(N.1) A+0= 0+A=A(N.2) A0= 0;0A= 0


2.2 Rank of a Matrix

Let A be anm×n matrix. The rank of the set of row vectors (column vectors) ofA is calledtherow rank (column rank) of A. We note that the row and column ranks of a matrix are,in fact, equal:

Theorem 3. (Rank Theorem)

For any m×n matrix A, the row rank and the column rank are equal.

In view of the rank theorem, we will, henceforth, simply refer to therankof A [denotedr(A)].

If A is anm×n matrix, andB is ann× r matrix, then the following properties can beestablished:

(R.1) r(A) ≤min(m,n)(R.2) r(AB) ≤min(r(A), r(B)).If A is anm×m matrix, thenA is callednon-singularif the rank ofA is m. A is called

singular if the rank ofA is less thanm.

2.3 Inverse of a Matrix

Let A be anm×mmatrix. If B is anm×mmatrix satisfying

AB=BA= I (2.1)

thenA is calledinvertibleandB is called theinverseof A (denoted byA−1). The followingproperties can be established regarding inverses of matrices. [HereA andB arem×minvertible matrices].

(IN.1) (A−1)−1 =A(IN.2) (AB)−1 =B−1A−1

(IN.3) (A′)−1 = (A−1)′

2.4 Relationship Between Invertible and Non-Singular Ma-trices

Consider anm×m non-singular matrix,A. Then them column vectors ofA are linearlyindependent and is therefore a basis ofRm. By the basis theorem, given anyb ∈Rm, b can


be expressed as a linear combination of them column vectors. That is, for eachb ∈ Rm,there isx ∈Rm such that

Ax= b (2.2)

Applying this to each of the column unit vectorsei (i = 1, ...,m) in turn, we will get vectorsxi (i = 1, ...,m) such that

Axi = ei (2.3)

Defining a matrixX with xi representing the ith column, we get

AX = I (2.4)

Them row vectors ofA are linearly independent and is therefore a basis ofRm. Bythe basis theorem, given anyc ∈Rm, c can be expressed as a linear combination of themrow vectors. That is, for eachc ∈Rm there isy ∈Rm such that

yA= c (2.5)

Applying this to each of the row unit vectorsei (i = 1, ...,m) in turn, we will get vectorsyi (i = 1, ...,m) such that

yiA= ei (2.6)

Defining a matrixY with yi representing its ith row, we get

YA= I (2.7)

Using (2.4) and (2.7), we get

Y =Y(AX) = (YA)X =X (2.8)

Thus, we have a matrixX such that

AX =XA= I

which proves thatA is invertible, andX is the inverse ofA.Conversely, consider anm×m invertiblematrix,A. Then, there is anm×m matrix,B,

such that

AB=BA= I (2.9)


We claim that the column vectors ofA are linearly independent. For if they are linearlydependent, there is some non-zero vectorx such that

Ax= 0 (2.10)

Multiplying (2.10) byB we get

0=B(Ax) = (BA)x= Ix = x (2.11)

which is a contradiction. Thus, them column vectors ofA are linearly independent, andthe rank ofA is m. Thus,A is a non-singular matrix.



Problem 5 (Matrix Multiplication).

SupposeA is anm×n matrix,B is ann× r matrix, andC is anr ×smatrix. Verify that:

A(BC) = (AB)CSolution.

A=

⎡⎢⎢⎢⎢⎢⎣a11 ⋯ a1n⋮ ⋱ ⋮am1 ⋯ amn

⎤⎥⎥⎥⎥⎥⎦, B=

⎡⎢⎢⎢⎢⎢⎣b11 ⋯ b1r⋮ ⋱ ⋮bn1 ⋯ bnr

⎤⎥⎥⎥⎥⎥⎦, C=

⎡⎢⎢⎢⎢⎢⎣c11 ⋯ c1s⋮ ⋱ ⋮cr1 ⋯ crs

⎤⎥⎥⎥⎥⎥⎦

A(BC) =⎡⎢⎢⎢⎢⎢⎣

a11 ⋯ a1n⋮ ⋱ ⋮am1 ⋯ amn

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣(b11c11+⋅ ⋅ ⋅+b1rcr1) ⋯ (b11c1s+⋅ ⋅ ⋅+b1rcrs)⋮ ⋱ ⋮(bn1c11+⋅ ⋅ ⋅+bnrcr1) ⋯ (bn1c1s+⋅ ⋅ ⋅+bnrcrs)

⎤⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎣∑n

j=1a1 j(∑rk=1b jkck1) ⋯ ∑n

j=1a1 j(∑rk=1b jkcks)⋮ ⋱ ⋮

∑nj=1am j(∑r

k=1b jkck1) ⋯ ∑nj=1am j(∑r

k=1b jkcks)⎤⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎣∑r

k=1ck1(∑nj=1a1 jb jk) ⋯ ∑r

k=1cks(∑nj=1a1 jb jk)⋮ ⋱ ⋮

∑rk=1ck1(∑n

j=1am jb jk) ⋯ ∑rk=1cks(∑n

j=1am jb jk)⎤⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎣(a11b11+⋅ ⋅ ⋅+a1nbn1) ⋯ (a11b1r +⋅ ⋅ ⋅+a1nbnr)⋮ ⋱ ⋮(am1b11+⋅ ⋅ ⋅+amnbn1) ⋯ (am1b1r +⋅ ⋅ ⋅+amnbnr)

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣c11 ⋯ c1s⋮ ⋱ ⋮cr1 ⋯ crs

⎤⎥⎥⎥⎥⎥⎦= (AB)C

Problem 6 (Transpose of a Matrix).

SupposeA is anm×n matrix,B is ann× r matrix. Verify that:

(AB)′ =B′A′

Solution.

A=

⎡⎢⎢⎢⎢⎢⎣A1⋮Am

⎤⎥⎥⎥⎥⎥⎦, B=

⎡⎢⎢⎢⎢⎢⎣B1 ⋯ Br

⎤⎥⎥⎥⎥⎥⎦


Recall that the inner product has some nice properties: for two vectorsx andy of the samesize, we havexy= yx andx′y= xy′ = x′y′ = xy. Then we can write

(AB)′ = ⎛⎜⎝⎡⎢⎢⎢⎢⎢⎣

A1⋮Am

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣B1 ⋯ Br

⎤⎥⎥⎥⎥⎥⎦⎞⎟⎠′

=

⎡⎢⎢⎢⎢⎢⎣A1B1 ⋯ A1Br

⋮ ⋱ ⋮AmB1 ⋯ AmBr

⎤⎥⎥⎥⎥⎥⎦

′

=

⎡⎢⎢⎢⎢⎢⎣A1B1 ⋯ AmB1

⋮ ⋱ ⋮A1Br ⋯ AmBr

⎤⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎣B1A1 ⋯ B1Am⋮ ⋱ ⋮BrA1 ⋯ BrAm

⎤⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎣(B1)′(A1)′ ⋯ (B1)′(Am)′⋮ ⋱ ⋮(Br)′(A1)′ ⋯ (Br)′(Am)′

⎤⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎣(B1)′⋮(Br)′

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣(A1)′ ⋯ (Am)′

⎤⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎣B1 ⋯ Br

⎤⎥⎥⎥⎥⎥⎦

′⎡⎢⎢⎢⎢⎢⎣A1⋮Am

⎤⎥⎥⎥⎥⎥⎦

′

=B′A′

Problem 7 (Rank of a Matrix).

Let A be anm×n matrix, and suppose then column vectors ofA are linearly indepen-dent.

(a) Show thatm≥ n.(b) Show that the row rank ofA≥ n.[Do not use the Rank Theorem to answer any of the parts of the problem. You can, of

course, use any result, which appears in the lecture notes before the statement of the RankTheorem].

Solution.


(a) We are given thatA1, . . . ,An are linearly independent vectors inRm. If m<n we have

a violation of Corollary 1 in Chapter 1. Therefore it must be that m≥ n.

(b) We will show row rank(A) ≥ n by contradiction. Suppose row rank(A) = r < n≤m.Then the row vectorsA1, . . . ,Ar , each of size 1×n, are a basis forA1, . . . ,Am. [IfusingA1, . . . ,Ar seems troublesome, remember that we can reorder the rows ofAwithout changing the rank.]

Consider the system ofr equations

λ1a11+⋅ ⋅ ⋅+λna1n = 0

⋮λ1ar1+⋅ ⋅ ⋅+λnarn = 0

whereai j is the jth element of theith row vectorAi. If we defineλ = [λ1⋯ λn]′ thenthis can be written more compactly as

⎡⎢⎢⎢⎢⎢⎣A1⋮Ar

⎤⎥⎥⎥⎥⎥⎦λ = 0.

This is a system ofr homogeneous linear equations in then> r unknownsλ1, . . . ,λn,so by Corollary 2 in Chapter 1, the system has a solutionλ1, . . . ,λn not all zero.

Now, sinceA1, . . . ,Ar are a basis for the rows ofA, any rowAk can be expressed asa linear combination ofA1, . . . ,Ar . That is, there areµ1, . . . ,µr such that

Ak = µ1A1+⋅ ⋅ ⋅+µrAr .

Taking the inner product of both sides withλ, we have

Akλ = µ1(A1λ)+⋅ ⋅ ⋅+µr(Arλ)= µ1(0)+⋅ ⋅ ⋅+µr(0)= 0

This must hold for every rowk, so the following equations hold:

A1λ = λ1a11+⋅ ⋅ ⋅+λna1n = 0

⋮Amλ = λ1am1+⋅ ⋅ ⋅+λnamn= 0


This is the same as writingλ1A1+⋅ ⋅ ⋅+λnAn = 0

for λ1, . . . ,λn not all zero. That means the columns ofA are linearly dependent,which is a contradiction. So it must be that row rank(A) ≥ n.

Problem 8 (Singular and Non-Singular Matrices).

(a) LetA be anm×1 matrix and letB be a 1×m matrix, wherem≥ 2. Let C be them×mmatrix, defined byC=AB. Show thatC must be a singular matrix.

(b) Let A be anm×n matrix and letB be ann×m matrix. LetC be them×m matrix,defined byC=AB. If n<m, canC be non-singular ? Explain your answer carefully.

Solution.

(a) Let

A=

⎡⎢⎢⎢⎢⎢⎣a11⋮am1

⎤⎥⎥⎥⎥⎥⎦, B= [b11 ⋯ b1m] , C=

⎡⎢⎢⎢⎢⎢⎣a11b11 ⋯ a11b1m⋮ ⋱ ⋮am1b11 ⋯ am1b1m

⎤⎥⎥⎥⎥⎥⎦To show thatC is singular, we want to show that rank(C) < m. Suppose to thecontrary that rank(C) =m. Then the columns ofC are linearly independent. Now, ifb1k =0 for somek ∈ {1, . . . ,m} thenCk =0 and we reach a contradiction immediately.So, supposeb1i ≠ 0 for all i = 1, . . . ,m. Consider the system of equations

λ1C1+λ2C2+⋅ ⋅ ⋅+λmCm= λ1b11

⎡⎢⎢⎢⎢⎢⎣a11⋮am1

⎤⎥⎥⎥⎥⎥⎦+λ2b12

⎡⎢⎢⎢⎢⎢⎣a11⋮am1

⎤⎥⎥⎥⎥⎥⎦+⋅ ⋅ ⋅+λmb1m

⎡⎢⎢⎢⎢⎢⎣a11⋮am1

⎤⎥⎥⎥⎥⎥⎦= 0

Choosingλ1 =1

b11≠ 0,λ2 = − 1

b12≠ 0, andλ j = 0 for j = 3, . . . ,m, the above equations

hold. This means that the columns ofC are linearly dependent, which is a contra-diction. So we conclude that rank(C) <m, which meansC is singular.

(b) We will first prove result (R.2) from Chapter 2:

rank(AB) ≤min{rank(A),rank(B)}This result will make it very easy to show thatC cannot be non-singular whenn<m.

We get (R.2) by showing both rank(AB) ≤ rank(A) and rank(AB) ≤ rank(B). Let’sprove the first statement and hope the second one follows similarly. Consider anylinear combination of the columns ofAB: if y is such a linear combination, then


there exists somem×1 vectorx such thaty= (AB)x. Since(AB)x= A(Bx), we cansay thaty is also a linear combination of the columns ofA, using the vectorBx asthe weights.

[Note that we cannot say the reverse. Ify′ is a linear combination of the columns ofA, then there is somez′ such thaty′ =Az′. How do we fitB in on the right hand side?If m= n andB is invertible, we can writey′ = (AB)(B−1z′) = Az′, so that the vectorB−1z′ provides the weights. But for this problem we are assumingn<m.]

Seeking contradiction, suppose rank(AB) = r > q= rank(A). Now, if (AB)k denotesthekth column ofAB, then some linearly independent vectors(AB)1, . . . ,(AB)r forma basis for the columns ofAB. [If using (AB)1, . . . ,(AB)r seems troublesome, re-member that we can reorder the columns ofAB without changing the rank.] Sim-ilarly, some linearly independent column vectorsA1

, . . . ,Aq form a basis for thecolumns ofA.

Since every column ofAB is trivially a linear combination of the columns ofAB, it isalso a linear combination of the columns ofA. In particular, each(AB)1, . . . ,(AB)ris a linear combination ofA1

, . . . ,An. Moreover, since eachA1, . . . ,An is a linear

combination ofA1, . . . ,Aq, we have that each(AB)1, . . . ,(AB)r is a linear combina-

tion of A1, . . . ,Aq. Now, sincer > q we have by the Fundamental Theorem on Vector

Spaces that the(AB)1, . . . ,(AB)r are linearly dependent. This is a contradiction, sowe conclude that rank(AB) ≤ rank(A).A similar argument establishes that rank(B′A′) ≤ rank(B′). Now, note that for anymatrix D,

rank(D) = row rank(D) = col rank(D′) = rank(D′).Therefore we have that

rank(AB) = rank((AB)′) ≤ rank(B′) = rank(B).With our finding above that rank(AB) ≤ rank(A), we have now proved that

rank(AB) ≤min{rank(A),rank(B)},which is result (R.2).

By result (R.1) in Chapter 2, we have rank(A) ≤ min{m,n} = n and rank(B) ≤min{n,m} = n. Then we only have to apply result (R.2) to see that rank(C) =rank(AB) ≤ n<m, which shows thatC must be singular.

Problem 9 (Inverse of a Matrix).


Let A be ann×n matrix, which satisfies:

ai j = { 1 for all i, j ∈ {1, ...,n} with j ≤ i0 otherwise

Show thatA has an inverse.[Do not use your computer to obtain the inverse matrix].

Solution.

A=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 ⋯ 0 01 1 0 ⋯ 0 01 1 1 ⋯ 0 0⋮ ⋮ ⋮ ⋱ ⋮ ⋮1 1 1 ⋯ 1 01 1 1 ⋯ 1 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦We know that

A has an inverse⇐⇒ A is non-singular

⇐⇒ rank(A) = n

⇐⇒ the columns ofA are linearly independent

Therefore, to showA has an inverse we only need to show that the columns ofA are linearlyindependent. Suppose to the contrary thatA1

, . . . ,An are linearly dependent, so that thereexistλ1, . . . ,λn not all zero satisfying

λ1A1+⋅ ⋅ ⋅+λnAn =

⎡⎢⎢⎢⎢⎢⎢⎢⎣

λ1

λ1+λ2⋮λ1+⋅ ⋅ ⋅+λn

⎤⎥⎥⎥⎥⎥⎥⎥⎦= 0

This implies thatλi = 0 for all i = 1, . . . ,n, which is a contradiction. So we conclude thatthe columns ofA are linearly independent, which shows thatA has an inverse as outlinedabove.


References:This material on matrix algebra can be found in standard texts like F. Hohn: Elemen-

tary Matrix Algebra(Chapters 1 and 6) andG. Hadley: Linear Algebra(Chapter 3). Agood discussion on the rank of a matrix is inD. Gale: Theory of Linear Economic Models(Chapter 2), and a proof of the “rank theorem” can be found there. You will also find agood exposition of matrix algebra inR. Dorfman, P.A. Samuelson and R.M. Solow: LinearProgramming and Economic Analysis(Appendix B).

Chapter 3

Simultaneous Linear Equations

3.1 System of Linear Equations

Consider a system ofm linear equations inn unknowns, written as

a11x1+ ...+a1nxn = b1

−−−−−−−−− (*)

am1x1+ ...+amnxn = bm

In matrix-vector notation, we can write this as

Ax= b

whereA is anm×n matrix, b is a (column) vector inRm (or anm×1 matrix) andx is a(column) vector inRn (or ann×1 matrix).

In analyzing a system of linear equations like (*), the following questions naturallyarise:

(i) (Existence) Does there exist a solution to (*)? [Are the equations “consistent”?](ii) (Uniqueness) If there exists a solution to (*), is it unique? [Are the equations

“determinate”?](iii) (Computation) If there exists a solution to (*), how canwe find such a solution?

3.2 Existence of Solutions

The system of equations (*) is calledhomogeneousif b= 0, andnon-homogeneousif b≠ 0.

28

CHAPTER 3. SIMULTANEOUS LINEAR EQUATIONS 29

If the system is homogeneous, there is always a trivial solution, namelyx= 0. If thesystem is non-homogeneous then, in general, there need not exist a solution to (*). Forinstance consider the system:

(i) 2x1+4x2 = 5(ii) x1+2x2 = 1Notice that if we multiply equation (ii) by 2, we get

2x1+4x2 = 2

wich contradicts(is inconsistent with) the equation (i). So, there does not exist a solutionto the system of equations given by (i) and (ii): the equations (i) and (ii) are inconsistent.

In general, if we look at a system of equations like (*), we would like to tell, givenAandb, whether there is a solution to (*).

We start our discussion of the existence of solutions to linear equations by noting asimple consequence of the “rank theorem”.

Proposition 1. Suppose a1, ...,an are linearly independent vectors inRm, and b is inRn,then there is a vector y inRm such that

yai = bi i = 1, ...,n

Proof. Let A be them×n matrix with columnsa1, ...,an. By the rank theorem, the rows of

A have rankn. Let A1, ...,An be a row basis; then we haven linearly independent vectorsin Rn, and hence by the basis theorem,b is a linear combination ofA1, ...,An. That is,there exist numbersλ1, ...,λn such that

n∑i=1

λiAi = b

Definey in Rm to be(λ1, ...,λn,0, ...,0). Then

m∑i=1

yiAi = b, or yA= b

Consequentlyyai = bi for i = 1, ...,n sinceai =Ai, the ith column ofA.Using the above result, one can provide the following criterion for the solvability of

linear equations.


Theorem 4. Let A be an m×n matrix, and let c be inRm. Then exactly one of the followingalternatives holds. Either the equation

Ax= c (3.1)

has a solution, or the equations

yA= 0, yc= 1 (3.2)

have a solution.

Proof. Suppose (3.1) has no solution. LetA1, ...,Ar be a column basis ofA. Then

A1, ...,Ar

,c is a linearly independent set of vectors. Otherwise,c can be expressed as alinear combination ofA1

, ...,Ar and therefore ofA1, ...,An. This contradicts our hypothesis

that (3.1) has no solution.Defineb in Rr+1 by b= (0, ...,0,1). Then by Proposition 1, there isy in Rm

, such that

yAi = 0 f or i = 1, ...,r, and yc= 1 (3.3)

SinceA1, ...,Ar is a column basis ofA, given anyAi(i = 1, ...,n),Ai can be expressed as a

linear combination ofA1, ...,Ar . So (3.3) implies thatyAi = 0 for i = 1, ...,n. Thus,y is a

solution to (3.2).On the other hand, if (3.1) has a solution (sayx), then (3.2) cannot have a solution.

For if it did (sayy), then multiplying (3.1) byy we get

1= yc= y(Ax) = (yA)x= 0

a contradiction.Finally, there is the criterion for solvability stated in terms of the rank of the relevant

matrices.Let A be anm×n matrix andc be a vector inRm. Then them×(n+1)matrix given by

Ac ≡ (A1, ...,An

,c) is known as theaugmented matrix.

Theorem 5. Let A be an m×n matrix and c be a vector inRm. Then the system ofequations

Ax= c (3.4)

has a solution if and only ifrankA= rankAc (3.5)


Proof. Let r be the rank ofA. We clearly haverank Ac ≥ rankA. SupposerankAc >

rankA. Then we can find(r +1) linearly independent vectors from the columns ofAc.Clearly this set must include the vectorc, otherwiserankA> r. Thus there is a set ofrcolumn vectors ofA, call themA1

, ...,Ar , which, together withc, form a set of linearlyindependent vectors. Thus,A1

, ...,Ar are linearly independent and so is a column basis ofA. If (3.4) has a solution, sayx, thenc is a linear combination ofA1

, ...,An and thereforeof A1

, ...,Ar . But then the set(A1, ...,Ar

,c) is linearly dependent, a contradiction. Thus if(3.5) does not hold, then (3.4) does not have a solution.

On the other hand, suppose (3.5) holds. Thus ifA1, ...,Ar is a column basis ofA, then(A1

, ...,Ar,c) is a linearly dependent set. Since(A1

, ...,Ar) is a linearly independent set,c can be expressed as a linear combination of the vectors(A1

, ...,Ar) and therefore of thevectors(A1

, ...,An). So (3.4) has a solution.

3.3 Uniqueness of Solutions

Theorem 6. Let A be an m×n matrix and let c be a vector inRm. Then the system ofequations

Ax= c (3.6)

has a unique solution if and only if

rankA= rankAc = n (3.7)

Proof. If (3.7) holds, then by Theorem 5, there is a solution to (3.6). Suppose, contrary tothe assertion of the Theorem, there arex1

, x2 in Rn, x1 ≠ x2 which both solve (3.6). Then

A(x1−x2) = 0

Thus, the column vectors ofA are linearly dependent [since(x1−x2) ≠ 0]. So therankA<n, a contradiction.

Suppose, next, that (3.7) does not hold. IfrankA≠ rankAc, there is no solution to(3.6) by Theorem 5, and we are done. IfrankA= rankAc, and (3.7) is violated, then wehaverankA= rankAc < n. It follows that (a) there is a solution,x, to (3.6); and (b) thencolumn vectors ofA, namelyA1

, ...,An are linearly dependent. Using (b), there isy ∈Rn,y≠ 0 such that

Ay= 0 (3.8)

But, then, clearly(x+y) also solves (3.6), and(x+y) ≠ x (sincey≠ 0), so (3.6) does nothave a unique solution.


3.4 Calculation of Solutions

The most important case to be considered in the actual calculation of solutions is the caseof n linear equations inn unknowns. LetA be ann×n matrix, andc be a vector inRn.We have the system of equations given by

Ax= c (3.9)

If (3.9) has a unique solution, then we have noted thatrankA=n. Conversely ifrankA=n,then A1

, ...,An is a basis ofRn, and soc can be expressed as a linear combination ofA1

, ...,An, yielding a solution to (3.9). Furthermore, such a solutionmust be unique sincerankA= rankAc = n.

We consider, therefore, in what follows how to calculatethe solution to (3.9) whenrankA= n. SincerankA= n, A is a non-singular matrix. It follows that it has ann×ninverse matrix (denoted byA−1) such that

A−1A=AA−1 = I (3.10)

Pre-multiplying (3.9) byA−1 and using (3.10), we obtain

x=A−1c (3.11)

asthesolution to (3.9).In terms of calculating this solution, then, it remains to learn how to compute the

inverse of a non-singular matrix. This leads us naturally into the study of “determinants”.

3.5 Determinants

Let A be ann×n matrix. We can associate withA a number, denoted by∣A∣, called thedeterminant of A.

The determinant of then×n matrix, is defined recursively as follows.(1) For a 1×1 matrix, which is, of course, a number, we define the determinant to

be the number itself.(2) For anym×m(m≥ 2) matrix, thecofactor Ai j of the elementai j is (−1)i+ j

times the determinant of the submatrix obtained fromA by deleting rowi and columnj.Thedeterminantof them×mmatrix is then given by

∣A∣ = n∑j=1

a1 jA1 j


Thus using (2), and knowing (1), the determinant of a 2×2 matrix is:

a11a22−a12a21

This information can then be used in (2) again to obtain the determinant of a 3×3 matrix:

a11[a22a33−a32a23]−a12[a21a33−a31a23]+a13[a21a32−a31a22]This procedure can be continued to obtain the determinant ofanyn×n matrix.

It is implicit in the definition of∣A∣ that the “expansion” is done by the first row. How-ever, it can be shown that for everyi ∈ [1, ...,n],

∣A∣ = n∑j=1

ai j Ai j

so that expansion by any row will given the same result. Indeed, expansion by any columnwill also give the same result. That is, for everyj ∈ [1, ...,n],

∣A∣ = n∑i=1

ai j Ai j

The following properties of determinants can be established:

(D.1) ∣A∣ = ∣A′∣(D.2) The interchange of any two rows will alter the sign, butnot the numerical value, of

the determinant.

(D.3) The multiplication of any one row by a scalark will change the determinantk-fold.

(D.4) The addition of a multiple of any row to another row willleave the determinantunaltered.

(D.5) If one row is a multiple of another row, the determinantis zero.

(D.6) The expansion of a determinant by “alien” co-factors yields a value of zero. Thatis,

n∑j=1

ai j Ak j = 0 i f i ≠ k

[Here, the expansion is by theith row, using co-factors ofkth row].

(D.7) ∣AB∣ = ∣A∣ ∣B∣The above properties (D.2) - (D.5) hold if the word “row” is replaced uniformly by

“column” in each statement.


3.6 Matrix Inversion

Now, we get back to the problem of computing the inverse of a non-singular matrix. Wefirst note the following result.

Theorem 7. Let A be an n×n matrix. Then A is invertible if and only if∣A∣ ≠ 0. Further-more, in case A is invertible,∣A−1∣ = ∣A∣−1.

Proof. SupposeA is invertible. Then

A A−1 = I

so 1= ∣I ∣ = ∣AA−1∣ = ∣A∣ ∣A−1∣ using property (D.7) of determinants, noted above. Conse-quently∣A∣ ≠ 0, and∣A−1∣ = ∣A∣−1.

Suppose, next, thatA is not invertible. Then,A is singular and so one of its columns(say,A1) can be expressed as a linear combination of its other columnsA2

, ...,An. That is,

A1 =n∑

i=2λiAi

Consider the matrix,B, whose first column is[A1− n∑i=2

λiAi] and whose other columns are

the same as those ofA. Then, the first column ofB is zero, and so∣B∣ = 0. By property(D.4) of determinants (noted above),∣B∣ = ∣A∣, and so∣A∣ = 0.

For ann×n matrix,A, we define theco-factor matrix of Ato be then×n matrix givenby

C=

⎡⎢⎢⎢⎢⎢⎣A11 A12 ...... A1n⋮ ⋮ ⋮ ⋮An1 An2 ...... Ann

⎤⎥⎥⎥⎥⎥⎦The transpose ofC is called theadjoint of A, and denoted byad j A.

Now, by the rules of matrix multiplication,

AC′ =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

n∑j=1

a1 jA1 jn∑

j=1a1 jA2 j ......

n∑j=1

a1 jAn j

⋮ ⋮n∑

j=1an jA1 j

n∑j=1

an jA2 j ......

n∑j=1

an jAn j

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎣∣A∣ 0 ...... 0⋮0 0 ...... ∣A∣

⎤⎥⎥⎥⎥⎥⎦


This yields the equation

AC′ = ∣A∣ I (3.12)

If A is non-singular (that is invertible) then there isA−1 such that

AA−1 =A−1A= I (3.13)

Pre-multiplying (3.12) byA−1 and using (3.13),

C′ = ∣A∣A−1

SinceA is non-singular, we have∣A∣ ≠ 0, and

A−1 =C′∣A∣ = ad jA∣A∣ (3.14)

Thus (3.14) gives us a formula for computing the inverse of a non-singular matrix in termsof the determinant and cofactors ofA.

3.7 Cramer’s Rule

Recall that we wanted to calculatethe (unique) solution of a system ofn equations innunknowns given by

Ax= c (3.15)

whereA is ann×n matrix, andc is a vector inRn.To obtain a unique solution, we saw that we must haveA non-singular, which now

translaltes to the condition “∣A∣ ≠ 0”. The unique solution to (3.15) is then

x=A−1c=ad jA∣A∣ c (3.16)

Let us evaluatex1, using (3.16). This can be done by finding the inner product ofx withthe first unit vector,e1 = (1,0, ...,0). Thus,

x1 = e1x=e1ad jA∣A∣ c

=[A11A21 An1]c∣A∣


= [c1A11+c2A21+ .....+cnAn1]/ ∣A∣

=

RRRRRRRRRRRRRRc1 a12 .. a1n⋮cn an2 .. ann

RRRRRRRRRRRRRR∣A∣−1

This gives us an easy way to compute the solution ofx1. In general, in order to calculatexi, replace theith column ofA by the vectorc and find the determinant of this matrix.Dividing this number by the determinant ofA yields the solutionxi. This rule is known asCramer’s Rule.


3.8 Appendix: System of Homogeneous Linear Equations

Let A be ann×n matrix. Consider the following system of homogeneous linearequations:

Ax= 0 (1)

SupposeA is singular, withrank(A) = r, where 1≤ r < n. Let S= {x ∈Rn ∶Ax= 0}.(a) Show thatScontains some non-zero vector.(b) Letq be the rank ofS. Show that 1≤ q< n.(c) Let{x1

, ...,xq} be a set of vectors inS, which is linearly independent [the existenceof such a set of vectors is guaranteed by (b)]. Show that thereare vectorsyq+1

, ...,yn inRn, such that the set{x1

, ...,xq,yq+1

, ...,yn} is a basis ofRn. [Use the result of problem 3in problem set 2].

(d) LetT = {v ∈Rn ∶ v=Azfor somez∈Rn}. Show that anyv ∈ T can be expressed as alinear combination of{zq+1

, ...,zn}, wherezj =Ayj for j = q+1, ...,n.(e) Show that the set{zq+1

, ...,zn} is linearly independent, and therefore the rank ofTis equal to(n−q).

(f) Show that rank ofT is also equal tor.(g) Conclude that:

rank(S) = n− rank(A) (*)

(a) Since the rank ofA is less thann, then column vectors ofA are linearly dependent.Consequently, there is somec ∈Rn, with c≠ 0, such thatAc= 0. This c ∈S.

(b) Sincec ∈ S, with c≠ 0, the rank ofS is ≥ 1. SinceS is a subset ofRn, and the rankof Rn is n, the rank ofS is ≤ n. To show thatq < n, suppose on the contrary thatq = n.Then, we can findn vectorsx1

, ..,xn in S, such that{x1, ..,xn} is linearly independent, and

is therefore a basis ofRn.Sincer ≥ 1, we can find at least one column vector among the column vectors of A,

which is non-zero; letAk be such a column vector ofA. The k-th unit vectorek ∈Rn canbe written as a linear combination of the vectors in{x1

, ..,xn}:ek = λ1x1+⋯+λnxn (2)

Pre-multiplying (2) byA, we obtain:

Ak =Aek = λ1Ax1+⋯+λnAxn = 0

sincex1, ..,xn are inS. But, this contradicts the fact thatAk is non-zero. Thus, we must

haveq< n.


(c) Since the rank ofSis q, we can find a set of vectorsx1, ..,xq in S, such that{x1

, ..,xq}is linearly independent. Since the rank ofRn is n, we can find vectorsy1

, ...,yn in Rn, suchthatE = {y1

, ...,yn}, is a basis ofRn. Using the result of problem 3 of problem set 2, wecan then find(n−q) vectors fromE, such that these vectors, together with theq vectorsx1, ..,xq constitute a basis ofRn. Denoting the(n−q) vectors (without loss of generality)

from E so chosen byyq+1, ...,yn

, the set{x1, ...,xq

,yq+1, ...,yn} is a basis ofRn.

(d) Since{x1, ...,xq

,yq+1, ...,yn} is a basis ofRn, anyz∈Rn can be expressed as a linear

combination of the set of vectors{x1, ...,xq

,yq+1, ...,yn} :

z= λ1x1+⋯+λqxq+λq+1yq+1+⋯+λnyn (3)

Pre-multiplying (3) byA, we obtain:

Az= λ1Ax1+⋯+λqAxq+λq+1Ayq+1+⋯+λnAyn

= λq+1Ayq+1+⋯+λnAyn

= λq+1zq+1+⋯+λnzn

(4)

the second line of (4) following from the fact that the vectors x1, ..,xq belong toS, and

the third line of (4) following from the definition ofzj for j = q+1, ...,n. Since anyv ∈ Tis equal toAz for somez ∈ Rn, this shows that anyv ∈ T can be expressed as a linearcombination of{zq+1

, ...,zn}.(e) Suppose that{zq+1

, ...,zn} is linearly dependent. Then, there are numbersαq+1, ..,αn,not all equal to zero such that:

αq+1zq+1+⋯+αnzn = 0 (5)

Using the definition ofzj for j = q+1, ...,n, we have:

A(αq+1yq+1+⋯+αnyn) = 0 (6)

Denoting(αq+1yq+1+⋯+αnyn) by w, we observe that (6) implies thatw∈S. Since the rankof S is q, and{x1

, ..,xq} is linearly independent, we know that{x1, ..,xq} is a basis ofS.

Consequently,w can be expressed as a linear combination of the set of vectors{x1, ..,xq}:

w= β1x1+⋯+βqxq (7)

Using the definition ofw, and (7), we get:

αq+1yq+1+⋯+αnyn−β1x1−⋯−βqxq = 0.

But this means that{x1, ...,xq

,yq+1, ...,yn} is linearly dependent, a contradiction. Thus,{zq+1

, ...,zn}must be linearly independent.


The vectors in{zq+1, ...,zn} belong (by definition) toT. Using (d) and the linear in-

dependence of{zq+1, ...,zn}, we infer that{zq+1

, ...,zn} is a basis ofT. Consequently, therank ofT is (n−q).

(f) Since the rank ofA is r, we can find a set ofr linearly independent vectors amongthen column vectors ofA. Without loss of generality, we let this set be{A1

, ...,Ar}. NotethatAi = Aei for i = 1, ...,r, so the vectors in{A1

, ...,Ar} belong toT. Since anyv ∈ T canbe expressed asAz for somez∈Rn, anyv ∈ T can be expressed as a linear combination ofthe vectors inU = {A1

, ...,An}.The rank ofU is r, and so{A1

, ...,Ar} is a basis ofU . Thus, anyA j ∈U can be expressedas a linear combination of the set of vectors in{A1

, ...,Ar}.It follows that anyv ∈ T can be expressed as a linear combination of the set of vectors

in {A1, ...,Ar}. Since{A1

, ...,Ar} is linearly independent,{A1, ...,Ar} is a basis ofT. Thus,

the rank ofT is r.(g) Using (e) and (f), we haver = n−q. This establishes (*).



Problem 10(System of Linear Equations: Existence and Uniqueness of Solutions).

Consider the following system of linear equations:

2x1+4x2 = 83x1+3x2 = 92x1+3x2 = 7

(1)

(a) Show, using the existence criterion discussed in class,that the system of equations(1) has a solution.

(b) Does the system of equations (1) have a unique solution ? Explain.

Solution.

We can write the system (1) asAx= b, where

A=

⎡⎢⎢⎢⎢⎢⎣2 43 32 3

⎤⎥⎥⎥⎥⎥⎦, x= [ x1

x2] , b=

⎡⎢⎢⎢⎢⎢⎣897

⎤⎥⎥⎥⎥⎥⎦, Ab =

⎡⎢⎢⎢⎢⎢⎣2 4 83 3 92 3 7

⎤⎥⎥⎥⎥⎥⎦(a) To show thatAx= b has a solution, we must show that rank(A) = rank(Ab). First,

rank(A) = 2: rank(A) ≤ 2 becauseA has only two columns, and rank(A) ≥ 2 becauseA1

,A2 are linearly independent. Second, rank(Ab) = 2: rank(Ab) ≥ 2 sinceA1b = A1

andA2b = A2 are linearly independent, and rank(Ab) < 3 sinceA3

b = 2A1b+A2

b, so thethree columns ofAb are linearly dependent. Because rank(Ab) = rank(A) = 2, thesystemAx= b has a solution by the Existence Theorem from Chapter 3.

(b) We can apply the Uniqueness Theorem from Chapter 3, since we have m= 3,n= 2.From part (a), rank(A)= rank(Ab)=2=n, so the systemAx=b has a unique solution.

Problem 11(System of Linear Equations: Existence of Solutions).

Consider the following system of linear equations:

3x1+x2+x3 = tx1−x2+2x3 = 1− tx1+3x2−3x3 = 1+ t

(2)

wheret is a real number. For what values oft will the system of equations (2) have asolution ? Explain.


Solution.

[Note: You are free to just solve the system, but the method below may help you solvecertain other existence or uniqueness problems.] We can write the system (2) asAx= b,where

A=

⎡⎢⎢⎢⎢⎢⎣3 1 11 −1 21 3 −3

⎤⎥⎥⎥⎥⎥⎦, x=

⎡⎢⎢⎢⎢⎢⎣x1

x2

x3

⎤⎥⎥⎥⎥⎥⎦, b=

⎡⎢⎢⎢⎢⎢⎣t

1− t1+ t

⎤⎥⎥⎥⎥⎥⎦, Ab =

⎡⎢⎢⎢⎢⎢⎣3 1 1 t1 −1 2 1− t1 3 −3 1+ t

⎤⎥⎥⎥⎥⎥⎦By the Existence Theorem from Chapter 3, the systemAx= b will have a solution whenrank(A) = rank(Ab). Now, rank(A) = 2 becauseA1

,A2 are linearly independent, butA1 =

2A2+A3. So we want to find the values oft for which rank(Ab) = 2. SinceA1,A2 are

linearly independent, if rank(Ab) = 2 thenA1,A2 will form a basis for the columns ofAb.

In particular, we can write

⎡⎢⎢⎢⎢⎢⎣t

1− t1+ t

⎤⎥⎥⎥⎥⎥⎦= λ1

⎡⎢⎢⎢⎢⎢⎣311

⎤⎥⎥⎥⎥⎥⎦+λ2

⎡⎢⎢⎢⎢⎢⎣1−13

⎤⎥⎥⎥⎥⎥⎦This is a system of three equations. By adding the first and second equations, we have4λ1 = 1, or λ1 =

14. By subtracting the second equation from the first equation, then sub-

tracting twice the third equation, we have−4λ2 =−3, orλ2 =34. With these values ofλ1,λ2,

the second equation givest = 32. This is the only value oft for which the systemAx= b has

a solution.

Problem 12(System of Linear Equations: Uniqueness of Solution).

Let A be anm×n matrix and letb be a vector inRm. Consider the following system of

linear equations:Ax= b (3)

Suppose (3) has a unique solution forevery b∈Rm. Canmbe different fromn ? Explain.

Solution.

Using the Uniqueness Theorem from Chapter 3, we are given thatfor all b ∈ Rm,rank(A) = rank(Ab) = n. That means theA1

, . . . ,An ∈Rm are linearly independent and anyb ∈ Rm is a linear combination ofA1

, . . . ,An. By the Basis Theorem, then, theA1, . . . ,An

are a basis forRm. Since rank(Rm) =m, it must be thatm= n.

Problem 13 (System of Homogeneous Linear Equations: Existence and Uniqueness ofSolutions).


Let A be ann×n matrix. Consider the following system of homogeneous linearequa-tions:

Ax= 0 (4)

(a) SupposeA is non-singular. Show that there is a unique solution to the system ofequations (4).

(b) SupposeA is singular. Show that there are an infinite number of distinct solutionsto the system of equations (4).

Solution.

(a) In class, the following result was presented for ann×n matrix A: the systemAx= bhas a unique solution if and only if rank(A) = n. Since we are given thatA is non-singular, we know rank(A) = n and so the systemAx= 0 has a unique solution.

(b) Let rank(A) = k < n. ThenAb = [A 0] also has rankk. By the Existence Theorem inChapter 3, then, the systemAx= 0 has a solution, since rank(A) = rank(Ab) = k.

SinceAx= 0 has a unique solution if and only ifA is non-singular, it must be thatAx= 0 has more than one solution. Suppose vectorsx andx′ solveAx= 0, withx≠ x′.Considerx′′ = λx+ (1−λ)x′ for someλ ∈ (0,1). Note thatx′′ ≠ x andx′′ ≠ x′. NowAx′′ = λAx+(1−λ)Ax′ = 0, sox′′ solvesAx= 0. Lettingλ vary over (0, 1), we havean infinite number of distinct solutions toAx= 0.

Problem 14(Determinant of Upper Triangular Matrix).

Let A be ann×n matrix, withai j = 0 wheneveri > j.(a) Show that:

detA=∏ni=1aii

(b) Use (a) to verify thatA is non-singular if and only ifaii ≠ 0 for eachi ∈ {1, ...,n}.Solution.

A=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

a11 a12 ⋯ a1,n−1 a1n

0 a22 ⋯ a2,n−1 a2n⋮ ⋮ ⋱ ⋮ ⋮0 0 ⋯ an−1,n−1 an−1,n

0 0 ⋯ 0 ann

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦


(a) Let’s prove this by induction.

Base case: Letn=1. If B is a 1×1 matrix, then detB=b11=∏1i=1bii by the definition

of a determinant.

Inductive case: Letn> 1. Assume that for any(n−1)×(n−1)matrixC with ci j = 0for all i > j, we have detC =∏n−1

i=1 cii . Now consider anyn×n matrix A with ai j = 0for all i > j. Expanding by the last row, we have

detA= an1An1+⋅ ⋅ ⋅+annAnn

= ann(−1)n+n

RRRRRRRRRRRRRRRRRR

a11 a12 ⋯ a1,n−1

0 a22 ⋯ a2,n−1⋮ ⋮ ⋱ ⋮0 0 ⋯ an−1,n−1

RRRRRRRRRRRRRRRRRR= ann

n−1∏i=1

aii

=n∏

i=1aii

where the third equality follows from the inductive hypothesis.

(b) As an “if and only if” statement, this requires proofs in bothdirections.

[Note: You are free to cite Theorem 7, Chapter 3, the proof of which is contained inthe answer below, in order to write a shorter proof for this problem.]

Claim: If the upper triangular matrix A is non-singular, then aii ≠ 0 for alli = 1, . . . ,n.

Proof: LetA be non-singular. ThenA has an inverse,A−1. Since

1= detI = detA−1A= (detA−1)(detA),we know that detA ≠ 0. If aii = 0 for any i ∈ 1, . . . ,n, then by (a) we would havedetA= 0, a contradiction. So it must be thataii ≠ 0 for all i = 1, . . . ,n.

Claim: If A is upper triangular and aii ≠ 0 for all i = 1, . . . ,n, then A is non-singular.

Proof: Letaii ≠ 0 for all i = 1, . . . ,n. Then by (a), detA≠ 0. Seeking contradiction,


supposeA is singular. Without loss of generality, we can writeA1 =∑ni=2λiAi. Let

B=

⎡⎢⎢⎢⎢⎢⎣A1−∑n

i=2λiAi A2 ⋯ An

⎤⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎣0 A2 ⋯ An

⎤⎥⎥⎥⎥⎥⎦We know, by property (D.4) in Chapter 3, that detB= detA. But, expandingB bythe first column, we have detB= 0. This gives detA= 0, a contradiction. So we havethatA is non-singular.

Problem 15(Test of Linear Dependence of Vectors).

Let S= {x1,x2

, . . . ,xm} be a set of vectors inRn, and letG be them×mmatrix defined by:

G=

⎡⎢⎢⎢⎢⎢⎣x1x1 ⋯ x1xm

⋮ ⋯ ⋮xmx1 ⋯ xmxm

⎤⎥⎥⎥⎥⎥⎦wherexix j is the inner product ofxi andx j , for i = 1, . . . ,mand j = 1, . . . ,m.

Show thatS is linearly dependent if and only if:

detG= 0.

Solution.

As an “if and only if” statement, this requires proofs in bothdirections.

Claim: If S is linearly dependent, then detG= 0.

Proof: SupposeS is linearly dependent. Then, without loss of generality, wecan writex1 = λ2x2+⋅ ⋅ ⋅+λmxm. Substituting this intoG in a clever way, we have

G=

⎡⎢⎢⎢⎢⎢⎢⎢⎣

(λ2x2+⋅ ⋅ ⋅+λmxm)x1 ⋯ (λ2x2+⋅ ⋅ ⋅+λmxm)xm

x2x1 ⋯ x2xm

⋮ ⋱ ⋮xmx1 ⋯ xmxm

⎤⎥⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎣

λ2x2x1+⋅ ⋅ ⋅+λmxmx1 ⋯ λ2x2xm+⋅ ⋅ ⋅+λmxmxm

x2x1 ⋯ x2xm

⋮ ⋱ ⋮xmx1 ⋯ xmxm

⎤⎥⎥⎥⎥⎥⎥⎥⎦


That is, the rows ofG are linearly dependent:G1 = λ2G2+ ⋅ ⋅ ⋅ +λmGm. SoG is singular,which is the same as sayingG is not invertible, which is the same as detG= 0.

Claim: If det G= 0, then S is linearly dependent.

Proof: Suppose detG= 0. This is equivalent to sayingG is not invertible, which is equiv-alent to sayingG is singular. That is, the columns ofG are linearly dependent: there existλ1, . . . ,λm not all zero such that

λ1

⎡⎢⎢⎢⎢⎢⎣x1x1

⋮xmx1

⎤⎥⎥⎥⎥⎥⎦+⋅ ⋅ ⋅+λm

⎡⎢⎢⎢⎢⎢⎣x1xm

⋮xmxm

⎤⎥⎥⎥⎥⎥⎦= 0

We can write this as the system of equations

x1(λ1x1+⋅ ⋅ ⋅+λmxm) = 0

⋮xm(λ1x1+⋅ ⋅ ⋅+λmxm) = 0

Now, multiply theith equation byλi and sum all the equations for

(λ1x1+⋅ ⋅ ⋅+λmxm)(λ1x1+⋅ ⋅ ⋅+λmxm) = 0

Let y = λ1x1+ ⋅ ⋅ ⋅ +λmxm. By (I.4) in Chapter 1,yy= 0 if and only if y = 0. This meansλ1x1+⋅ ⋅ ⋅+λmxm= 0 for λ1, . . . ,λm not all zero. ThusS is linearly dependent.


References:This material is based onLinear Algebraby G. Hadley(Chapters 3, 5),Elementary

Matrix Algebraby F. Hohn(Chapters 2, 6, 7) andThe Theory of Linear Economic Modelsby D. Gale (Chapter 2). Some of this material is also covered inLinear Programming andEconomic Analysisby R. Dorfman, P.A. Samuelson and R.M. Solow(Appendix B).

Chapter 4

Characteristic Values and Vectors

4.1 The Characteristic Value Problem

Let C denote the set of complex numbers. Given ann×n real matrix, for whatnon-zerovectorsx ∈Cn, and for what complex numbersλ is it true that

Ax= λx (4.1)

This is known as thecharacteristic value problemor theeigenvalue problem.If x≠0 andλ satisfy equation (4.1), thenλ is called acharacteristic valueor eigenvalue

of A, andx is called acharacteristic vectoror eigenvectorof A.Clearly (4.1) holds if and only if

(A−λI)x= 0 (4.2)

But (4.2) is a homogeneous system ofn equations inn unknowns. It has a non-zerosolution forx if and only if (A−λI) is singular; that is, if and only if

∣A−λI ∣ = 0 (4.3)

This equation is called thecharacteristic equationof A. If we look at the expression

f (λ) ≡ ∣A−λI ∣ (4.4)

we note thatf is apolynomialin λ; it is called thecharacteristic polynomialof A.

Example: Consider the 2×2 matrixA given by

A= [ 2 11 2

]47

CHAPTER 4. CHARACTERISTIC VALUES AND VECTORS 48

Then equation (4.3) becomes

∣ 2−λ 11 2−λ ∣ (4.5)

So,(4−2λ+λ2)−1= 0, which yields

(1−λ)(3−λ) = 0

Thus, the characteristic roots areλ = 1 andλ = 3.Puttingλ = 1 in (4.2), we get

[ 1 11 1

] [ x1

x2] = [ 0

1]

which yieldsx1+x2 = 0

Thus the general solution of the characteristic vector corresponding to the characteristicroot λ = 1 is given by (x1, x2) = θ(1, −1) f orθ ≠ 0

Similarly, corresponding to the characteristic rootλ = 3, we have the characteristic vectorgiven by (x1, x2) = θ(1,1) f orθ ≠ 0.

In developing the basic results on the characteristic-value problem, we note that, ingeneral, the characteristic equation will haven roots in the complex plane (by the “Fun-damental Theorem of Algebra”), since it is a polynomial equation (in λ) of degreen. [Ofcourse some of these roots might be repeated]. In general, the corresponding eigenvectorswill also have their components in the complex plane.

4.2 Characteristic Values, Trace and Determinant of aMatrix

If A is ann×n matrix, thetraceof A, denoted bytr(A), is the number defined by

tr(A) = n∑i=1

aii

The following properties of the trace can be verified easily [HereA, B andC aren×nmatrices, andλ ∈R].


(TR.1) tr(A+B) = tr(A)+ tr(B)(TR.2) tr(λA) = λ tr(A)(TR.3) tr(AB) = tr(BA)(TR.4) tr(ABC) = tr(BCA) = tr(CAB)

Let A be ann×n matrix. The characteristic polynomial ofA, defined in (4.4) abovecan generally be written as

∣A−λI ∣ = (−λ)n+bn−1(−λ)n−1+ ....+b1(−λ)+b0 (4.6)

whereb0, ...,bn−1 are the coefficients of the polynomial which are determined by the coef-ficients of theA-matrix.

On the other hand, ifλ1, ...,λn are the eigenvalues ofA, then the characteristic equation(4.3) can be written as

0= (λ1−λ)(λ2−λ)....(λn−λ) (4.7)

Using (4.3), (4.6), and (4.7) and “comparing coefficients” we can conclude that

bn−1 = λ1+λ2+ ...+λn

and

b0 = λ1λ2...λn

Also, by looking at the terms in the characteristic polynomial of A which would involve(−λ)n−1, we can conclude that

bn−1 = a11+a22+ ...+ann

Finally, puttingλ = 0 in (4.6), we get

b0 = ∣A∣Thus we might note two interesting relationships between the characteristic values, the

trace and the determinant ofA:

trA =n∑

i=1λi

∣A∣ = n∏i=1

λi


4.3 Characteristic Values and Vectors of Symmetric Ma-trices

There is considerable simplification in the theory of characteristic values ifA is asymmet-ric matrix. In this case, it can be shown that all the roots of (4.3) arereal.

Theorem 8. Let A be a symmetric n×n matrix. Then all the characteristic values of A arereal.

Proof. Supposeλ is a complex characteristic value, with associated complexcharacteristicvector,x. Then we have

Ax= λx (4.8)

Definex∗ to be the complex conjugate ofx, andλ∗ to be the complex conjugate ofλ. Then

Ax∗ = λ∗x∗ (4.9)

Pre-multiply (4.8) by(x∗)′ and (4.9) byx′ to get

(x∗)′Ax= λ(x∗)′x (4.10)

x′Ax∗ = λ∗x′x∗ (4.11)

Subtracting (4.11) from (4.10)

(x∗)′Ax−x′Ax∗ = (λ−λ∗)x′x∗ (4.12)

since(x∗)′x= x′x∗. Also,

x′Ax∗ = (x′Ax∗)′ = (x∗)′A′x= (x∗)′Ax

sinceA′ =A (by symmetry). Thus (4.12) yields

(λ−λ∗)x′x∗ = 0 (4.13)

Sincex≠ 0, we know thatx′x∗ is real and positive. Hence (4.13) implies thatλ = λ∗, soλis real.

We will develop the theory of eigenvalues and eigenvectorsonly for symmetric matri-ces.

Notice that once the eigenvalues are real, the system of equations

(A−λI)x= 0


will yield a non-zero solutionx in Rn if and only if

∣A−λI ∣ = 0

So the eigenvectors corresponding to the eigenvalues ofA will also be real vectors.If x is an eigenvector corresponding to an eigenvalueλ, then so istx, wheret is any

non-zero scalar. A normalized eigenvectoris an eigenvector with (Euclidean) norm equalto 1.

4.4 Spectral Decomposition of Symmetric Matrices

An n×n matrixC is called anorthogonal matrixif it is invertible, and its inverse equals itstranspose; that isC′ =C−1.

Theorem 9. Suppose A is an n×n symmetric matrix with n distinct eigenvalues,λ1, ...,λn.If x1

, ...,xn are (normalized) eigenvectors corresponding to the eigenvaluesλ1, ...,λn, thenthe matrix B, such that Bi, the ith column of B, is the vector xi, is an orthogonal matrix.

Proof. Pick any two distinct indicesi and j (so i ≠ j). Then we have

Axi = λixi (4.14)

andAxj = λ jx j (4.15)

Multiplying (4.14) by(x j)′ we get

(x j)′Axi = λi(x j)′xi (4.16)

Multiplying (4.15) by(xi)′ we get

(xi)′Axj = λ j(xi)′x j (4.17)

Now, (xi)′Axj is a number, so its transpose is the same number. Thus

(xi)′Axj = ((xi)′Axj)′ = (x j)′A′xi

But A′ =A by symmetry ofA. So

(xi)′Axj = (x j)′Axi


Using this in (4.16) and (4.17), we get

(λi −λ j)xix j = 0

Since the eigenvalues ofA are distinct,λi ≠ λ j for i ≠ j. Thusxix j = 0, soxi is orthogonalto x j .

It follows from this thatx1, ...,xn are linearly independent. For if they were dependent,

there would existµ1, ...,µn, not all zero, such that

µ1x1+ ...+µnxn = 0 (4.18)

Without loss of generality considerµ1 ≠ 0. Then premultiplying (4.18) by(x1)′, we get

µ1∥x1∥2 = 0

since(x1)′x j = 0 for all j ≠ 1. Since∥x1∥2 = 1, we getµ1 = 0, a contradiction. Thusx1, ...,xn are linearly independent.

If B is the matrix such thatBi, the ith column ofB, is the vectorxi, thenB is invertible,since we have shown thatB is non-singular. Also,

BB′ = I (4.19)

sincexix j = 0 for i ≠ j, andxix j = 1 for i = j. Thus premultiplying (4.19) byB−1 we get

B′ =B−1

HenceB is an orthogonal matrix.

Theorem 10. (Spectral Decomposition)Suppose A is an n×n symmetric matrix with n distinct eigenvalues,λ1, ...,λn. If B

is the n×n matrix with Bi, the ith column of B, being a (normalized) eigenvector of Acorresponding to the eigenvalueλi of A(i = 1, ...,n), then

A=BLB′ (4.20)

where L is the diagonal matrix with the eigenvalues of A on its diagonal.

Proof. For eachi = 1, ...,n, we have

ABi = λiBi


This can be written in compact form as

AB=BL (4.21)

where then×n matrix L is defined by

L =

⎡⎢⎢⎢⎢⎢⎢⎢⎣

λ1 0 ... 00 λ2 ... 0... ... ... ...

0 0 ... λn

⎤⎥⎥⎥⎥⎥⎥⎥⎦We have noted in Theorem 9 thatB is an orthogonal matrix. So, post-multiplying (4.21)by B−1, we get

A=BLB−1

But sinceB is orthogonal, we haveB−1 =B′, and thus we arrive at (4.20).Remark: The formula (4.20) “decomposes” the matrixA into a matrixL consisting of itseigenvalues, and the matricesB, B′ which consist of its eigenvectors.

4.5 Quadratic Forms

Definition 1. Let A be a symmetric n×n matrix. Then

(a) A is negative semi-definiteif hAh≤ 0 for all h in Rn.

(b) A is negative definiteif hAh< 0 for all h in Rn, h≠ 0.

(c) A is positive semi-definiteif hAh≥ 0 for all h inRn.

(d) A is positive definiteif hAh> 0 for all h in Rn, h≠ 0.

Let us concentrate on definition(d). Notice that the relevant inequality must holdfor everyvectorh≠ 0 in Rn. This means that if we already know that a symmetricn×nmatrix A is positive-definite, then we should be able to infer some useful properties ofAquite easily. On the other hand, it also means that if we do notknow that a symmetricn×n matrix A is positive definite, definition(d) by itself will not be very easy to check todetermine whetherA is positive definite or not.

We illustrate the first observation by noting that if a symmetric n×n matrix, A, ispositive definite then we can infer that all its diagonal elements must be positive. To seethis, note that theith diagonal element,aii , can be expressed as


(ei)′Aei = aii

whereei is the ith unit vector ofRn. SinceA is positive definite andei ≠ 0 is in Rn,definition (d) tells us that the left-hand side of the above equation is positive. Thus theright-hand side is also positive.

The second observation leads one to explore convenient characterizations of quadraticforms from which it should be easy to check whether a givenn×n symmetric matrixA ispositive definite or not. We provide two such characterizations in the next two sections.

4.6 Characterization of Quadratic Forms

Let A be ann×n symmetric matrix, with distinct eigenvaluesλ1, ...,λn. We have seen thatwe can define a matrixB, such thatBi, theith column ofB, is a normalized eigenvector ofAcorresponding to the eigenvalueλi(i =1, ...,n). This matrix is orthogonal, and furthermore

B′AB= L (4.22)

whereL is a diagonal matrix with the eigenvaluesλ1, ...,λn on its diagonal.Let x be an arbitrary non-zero vector inRn. SinceB is an orthogonal matrix we can

define the vectory∈Rn by y=B−1x=B′x, so thatBy= x, andy′B′ = x′. Then premultiplying(4.22) byy′ and post-multiplying it byy, we get

y′B′ABy= y′Ly

which yields, by definition ofy,x′Ax= y′Ly (4.23)

The right hand side of (4.23) is

y′Ly=n∑

i=1λiy2

i (4.24)

wherey= (y1, ...,yn). Suppose all the eigenvalues ofA are positive. Then sincex≠ 0, bydefinition ofy, we havey≠ 0, and (4.24) is positive. So, by (4.23),x′Ax> 0 for eachx≠ 0,andA is positive definite.

Conversely, ify is an arbitrary vector inRn, we definex=Byand soy′B′ = x′, and notethat if y≠ 0, thenx≠ 0 sinceB is orthogonal. Thus repeating the above calculations, weget (4.23) and (4.24). Now, supposeA is positive definite. Then choosing in turny= ej ,the jth unit vector, it follows from (4.24) thaty′Ly= λ j and from (4.23) thatλ j = x′Ax> 0.Thus, all the eigenvalues ofA are positive.

In the same way, characterization of the other quadratic forms can be obtained in termsof the signs of the eigenvalues of the matrix. We summarize these results as follows:


(a) A is positive (negative) definite if and only if every eigenvalue of Ais positive (neg-ative).

(b) A is positive (negative) semi-definite if and only if every eigenvalue of A is non-negative (non-positive).

Examples:Consider the following matrices:

A= [ −1 00 0

] ;B= [ −1 11 −3 ] ;C= [ 0 0

0 1]

The eigenvalues ofA are−1 and 0. SoA is negative semi-definite. The eigenvalues ofB are(−2+√2) and(−2−√2), which are both negative. SoB is negative definite. Theeigenvalues ofC are 0 and 1. SoC is positive semidefinite.

4.7 Alternative Characterization of Quadratic Forms

There is an alternative way to characterize quadratic formsin terms of the signs of the“principal minors” of the corresponding matrix.

If A is an n×n matrix, aprincipal minor of order r is the determinant of ther × rsubmatrix that remains when(n− r) rows and(n− r) columnswith the same indicesaredeleted fromA.Examples: If A is a 3×3 matrix, then the principal minors of order 2 are

∣ a11 a12

a21 a22∣ ; ∣ a11 a13

a31 a33∣ ; ∣ a22 a23

a32 a33∣

And, the principal minors of order 1 are

a11 ; a22 ; a33

The principal minor of order 3 is the determinant of the matrix.If A is ann×n matrix, theleading principal minor of order ris defined as

Dr =

RRRRRRRRRRRRRRa11 ⋯ a1r⋮ar1 ⋯ arr

RRRRRRRRRRRRRRThusD1 = a11; D2 = ∣ a11 a12

a21 a22∣ ;etc.

We note below the characterization result for positive and negative definite matrices.Let A be ann×n symmetric matrix.


(1’) (i) A is a positive definite if and only if all the leading principal minors of A arepositive.

(ii) A is a negative definite if and only if the leading principal minors of A alternatein sign, starting with negative[That is, therth leading principal minor,Dr , (wherer = 1, ...,n) has the same sign as(−1)r ].

Checking that a matrix is positive or negative semi-definite is somewhat more involved.The relevant results are stated below:

(2’) (i) A is positive semi-definite if and only if every principal minor of A of every orderis non-negative.

(ii) A is negative semi-definite if and only if every principal minor of A of odd orderis non-positive and every principal minor of even order is non-negative.

Examples:We re-examine the matrices which we studied above:

A= [ −1 00 0

] ;B= [ −1 11 −3 ] ;C= [ 0 0

0 1]

All the principal minors ofA of odd order are non-positive[a11 < 0 anda22 = 0], andthe (only) principal minor ofA of even order is non-negative[a11a22−a21a12= 0]. SoA isnegative semidefinite.

For the matrixB, the leading principal minor of order 1 is negative[b11= −1< 0], andthe leading principal minor of order 2 is positive[b11b22−b12b21= 2> 0]. SoB is negativedefinite.

The principal minors ofC of odd order are non-negative[c11 = 0 andc22 > 0]. Theprincipal minor ofC of even order is non-negative[c11c22−c12c21 = 0]. SoC is positivesemi-definite.


4.8 Appendix: Spectral Decomposition of Non-SymmetricMatrices

The characteristic roots of non-symmetric matrices need not be real numbers. However,when they are real numbers, one can develop a theory of spectral decomposition of non-symmetric matrices, using the methods already developed inthis course.

The following problem, split up into five parts, provides thesteps involved in such atheory. You might wish to work through the steps; it is entirely optional.

Let A be ann×n matrix (not necessarily symmetric), whose characteristicroots arereal and distinct; denote the roots byλ1, ...,λn.

(a) Show that for eachi ∈ {1, ...,n}, there exists a vectorxi ∈Rn, such thatxi ≠ 0, and:

(A−λiI)xi = 0

(b) The principal difference from the case in whichA is symmetric is that the char-acteristic vectorsx1

, ...,xn need not be orthogonal. To see this, consider the followingexample:

A= [ 2 43 1

](i) Show thatA has two real characteristic roots, which are distinct; callthemλ1 and

λ2.(ii) Obtain characteristic vectorsx1 ∈ R2 andx2 ∈ R2, corresponding toλ1 andλ2 re-

spectively.(iii) Show thatx1 is not orthogonal tox2.(c) Continuing now with the result obtained in (a) above, showthat the set{x1

, ...,xn}is linearly independent. Denote byX the matrix whose i-th column is given byxi fori ∈ {1, ..,n}. Note thatX has an inverse.

(d) Show that:AX =XΛ

whereΛ is a diagonal matrix with the characteristic rootsλ1, ...,λn on its diagonal (in thatorder). Note then that:

A=XΛX−1

(e) Show that fort = 1,2,3, ..., we have:

At =XL(t)X−1


whereL(t) is a diagonal matrix, withλt1, ...,λt

n on its diagonal (in that order).Remark:The above theory also goes through when the characteristic roots are complex, but

distinct. However, the methods developed in this course will not suffice to cover this case,because concepts such as linear independence and matrix inverse were developed in thiscourse starting with vectors inRn (notCn).



Problem 16(Non-Symmetric Matrices and Real Eigenvalues).

Let A be the 2×2 matrix, given by:

A= [ a11 a12

a21 a22]

wherea12≠ a21. Let p be the trace ofA, and letq be the determinant ofA. Assume that:

p> 0,q> 0, p2 > 4qand(p−q) < 1

Denote the characteristic roots ofA by λ1 andλ2.

(i) Show thatλ1 andλ2 are real and positive.(ii) Show that exactly one of the following alternatives must occur: (A) λ1 < 1 and

λ2 < 1; (B) λ1 > 1 andλ2 > 1.

Solution.

(a) Note thatp= tr(A) = a11+a22 andq= detA= a11a22−a12a21. Then we want to solve

0= f (λ) = det(A−λI) = (a11−λ)(a22−λ)−a12a21

= λ2−(a11+a22)λ+(a11a22−a12a21)= λ2− pλ+q

By the quadratic formula, the roots are

λ =p±√p2−4q

2

We are given thatp2 > 4q, so there are two distinct real roots. Without loss ofgenerality, let’s call the larger rootλ1. Then we have

λ1 =p+√p2−4q

2, λ2 =

p−√p2−4q

2, λ1 > λ2

Sincep> 0, we haveλ1 > 0. The smaller rootλ2 is positive if and only if

p−√p2−4q> 0

⇐⇒ p>√

p2−4q

⇐⇒ p2 > p2−4q

⇐⇒ q> 0


wherep>√

p2−4q ⇐⇒ p2 > p2−4q holds sincep> 0. Because we are givenq> 0,we haveλ2 > 0.

(b) If λ1 = 1 or λ2 = 1, thenp−q= 1, which is a contradiction of the given informationthat p−q< 1. So, in the proof below we consider only strict inequalities.

(A): λ1 < 1 andλ2 < 1. (not A):λ1 > 1 or λ2 > 1.

(B): λ1 > 1 andλ2 > 1. (not B):λ1 < 1 or λ2 < 1.

We want to show that (A) implies (not B) and that (not A) implies(B). This willestablish that exactly one of (A) or (B) always occurs.

Claim: (A) implies (not B). That is, if λ1 < 1 and λ2 < 1, then λ1 < 1 or λ2 < 1.

Proof: By hypothesis,λ1 < 1. So the claim holds.

Claim: (not A) implies (B). That is, if λ1 > 1 or λ2 > 1, then λ1 > 1 and λ2 > 1.

Proof: There are two cases to consider: one in whichλ1 > 1 and one in whichλ2 > 1.We must show that in both cases,λ1 > 1 andλ2 > 1.

Case 1: Supposeλ1 > 1. Then it follows thatp−2> −√p2−4q. Seeking contradic-tion, assumeλ2 < 1. This impliesp−2<

√p2−4q. So we have

∣p−2∣ <√p2−4q

⇐⇒ p2−4p+4< p2−4q

⇐⇒ p−q> 1

This is a contradiction of the given information thatp−q < 1, so it must be thatλ2 > 1.

Case 2: Supposeλ2 > 1. Thenλ1 > λ2 > 1.

Sinceλ1 > 1 andλ2 > 1 in both cases, the claim holds.

Problem 17(Eigenvalues and Eigenvectors of Symmetric Matrices).

Let A = (ai j ) be a symmetric 2×2 matrix. We know that it has only real eigenvalues;denote these byλ1 andλ2.

(a) Show that there isb= (b1,b2) ∈R2, with b≠ 0, such that(A−λ1I)b= 0. This shows

that there is a real eigenvector corresponding to the eigenvalueλ1.

(b) Definey= (y1,y2) as follows:y1 = b1+ ib1,y2 = b2+ ib2. Is y also an eigenvector ofA (corresponding to the eigenvalueλ1)? Explain.

Solution.


(a) Sinceλ1 solves det(A−λ1I) = 0, we know that(A−λ1I) is singular. Then thereareb1 andb2 not both zero such thatb1(A−λ1I)1+b2(A−λ1I)2 = 0. Lettingb =(b1,b2)′, we have that(A−λ1I)b= 0, sob is an eigenvector ofA, corresponding tothe eigenvalueλ1.

(b) We can writey= b+ ib. Then

(A−λ1I)y= (A−λ1I)b+(A−λ1I)ib = 0+0i = 0.

sinceb is an eigenvector of A, corresponding to the eigenvalueλ1. Soy is an eigen-vector of A, corresponding to the eigenvalueλ1.

Problem 18(Application of Spectral Decomposition).

Let A be the 2×2 matrix defined as follows:

A= [ 2 11 2

](a) Obtain the characteristic values and corresponding normalized characteristic vec-

tors of this matrix.(b) Use the Spectral Decomposition Theorem to show that for any positive integern,

the matrixAn can be written as:

An = [ (3n/2)+(1/2) (3n/2)−(1/2)(3n/2)−(1/2) (3n/2)+(1/2) ]Solution.

(a) The eigenvalues of A are the roots of the characteristic polynomial

f (λ) = det(A−λI) = (2−λ)2−1

= λ2−4λ+3

= (λ−1)(λ−3)The roots areλ1 = 1,λ2 = 3.

A normalized eigenvectorb1 = (b11,b

12)′ corresponding toλ1 = 1 solves

⎡⎢⎢⎢⎢⎢⎣1 1

1 1

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣

b11

b12

⎤⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎣0

0

⎤⎥⎥⎥⎥⎥⎦


This system impliesb11+b1

2 = 0, so thatb1 = (b11,−b1

1)′. The normalization constraintis

1= ∥b1∥ =√(b11)2+(−b1

1)2 =√

2(b11)2

This is solved byb11 =

1√2, so a normalized eigenvector corresponding toλ1 = 1 is

b1 = ( 1√2,− 1√

2)′.

A normalized eigenvectorb2 = (b21,b


⎡⎢⎢⎢⎢⎢⎣−1 1

1 −1⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣

b21

b22

⎤⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎣0

0

⎤⎥⎥⎥⎥⎥⎦This system impliesb2

1 = b22, so thatb2 = (b2

1,b21)′. The normalization constraint is

1= ∥b2∥ =√(b21)2+(b2

1)2 =√

2(b21)2

This is solved byb21 =

1√2, so a normalized eigenvector corresponding toλ2 = 3 is

b2 = ( 1√2,

1√2)′.

(b) Let

B= [ b1 b2 ] =⎡⎢⎢⎢⎢⎢⎣

1√2

1√2

− 1√2

1√2

⎤⎥⎥⎥⎥⎥⎦, Λ = [ 1 0

0 3]

By the Spectral Decomposition Theorem,A=BΛB′. BecauseB′ =B−1, we have

An =BΛnB′

=

⎡⎢⎢⎢⎢⎢⎣1√2

1√2

− 1√2

1√2

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣

1n 0

0 3n

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣

1√2− 1√

21√2

1√2

⎤⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎣1√2

3n 1√2

− 1√2

3n 1√2

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣

1√2− 1√

21√2

1√2

⎤⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎣3n

2 + 12

3n

2 − 12

3n

2 − 12

3n

2 + 12

⎤⎥⎥⎥⎥⎥⎦


Problem 19(Symmetric Matrices with Repeated Characteristic Roots).

Let A be the 3×3 matrix, defined by:

A=

⎡⎢⎢⎢⎢⎢⎣3 0 00 2 10 1 2

⎤⎥⎥⎥⎥⎥⎦(a) Show that the characteristic values ofA are 3 and 1, but 3 is a repeated root. That

is, show that the characteristic polynomialf (λ) = det(A−λI) can be written as a productof three factors as follows:

f (λ) = (1−λ)(3−λ)(3−λ)(b) Show that(b1)′ = [0,1/√2,−1/√2] is a normalized characteristic vector corre-

sponding to the characteristic rootλ1 = 1.(c) Show that(b2)′ = [0,1/√2,1/√2] is a normalized characteristic vector correspond-

ing to the characteristic rootλ2 = 3.(d) Show that there is another normalized characteristic vector, b3

, corresponding tothe characteristic rootλ2 = 3, which is orthogonal to bothb1 andb2

.

(e) DefineB as the 3× 3 matrix which hasb1,b2

,b3 as its first, second and thirdcolumns. DefineΛ to be the diagonal matrix:

Λ =⎡⎢⎢⎢⎢⎢⎣

1 0 00 3 00 0 3

⎤⎥⎥⎥⎥⎥⎦Show thatA=BΛB′.

Solution.

(a) The characteristic polynomial is

f (λ) =RRRRRRRRRRRRRR

3−λ 0 00 2−λ 10 1 2−λ

RRRRRRRRRRRRRR= (3−λ)[(2−λ)2−1]= (3−λ)(λ2−4λ+3)= (3−λ)(3−λ)(1−λ)


(b) We have that∥b1∥ = 1 and

(A−λ1I)b1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

2 0 0

0 1 1

0 1 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0

1√2

− 1√2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0

1√2− 1√

21√2− 1√

2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦= 0

Sob1 is a normalized eigenvector corresponding to the eigenvalue λ1 = 1.

(c) We have that∥b2∥ = 1 and

(A−λ2I)b2 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0

0 −1 1

0 1 −1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0

1√2

1√2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0

1√2− 1√

21√2− 1√

2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦= 0

Sob2 is a normalized eigenvector corresponding to the eigenvalue λ2 = 3.

(d) A normalized eigenvectorb3 = (b31,b

32,b


(A−λ2I)b3 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0

0 −1 1

0 1 −1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

b31

b32

b32

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0

0

0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦This system impliesb3

2 = b33.

Any eigenvectorb3 satisfyingb32 = b3

3 is orthogonal tob1, so this condition gives usno information about the components ofb3.

We needb3 to be orthogonal tob2, so we require

0=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0

1√2

1√2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

b31

b32

b32

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦=

1√2

b32+ 1√

2b3

2 =2√2

b32

This impliesb32 = 0, so any eigenvectorb3 satisfyingb3

2 = b33 = 0 is orthogonal tob2.

We can chooseb3 = (1,0,0)′. This is a normalized eigenvector corresponding toλ2 = 3 that is orthogonal to bothb1 andb2.


(e) Collecting the normalized eigenvectors above, we have

B= [ b1 b2 b3 ] =⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 1

1√2

1√2

0

− 1√2

1√2

0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦Then

BΛB′ =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 1

1√2

1√2

0

− 1√2

1√2

0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0

0 3 0

0 0 3

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1√2− 1√

2

0 1√2

1√2

1 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 3

1√2

3√2

0

− 1√2

3√2

0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1√2− 1√

2

0 1√2

1√2

1 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎣3 0 00 2 10 1 2

⎤⎥⎥⎥⎥⎥⎦=A

Problem 20(Positive Definite Matrices).

Let A be ann×n symmetric matrix, which is positive definite. For eachk ∈ {1, . . . ,n},define the submatrixA(k) by:

A(k) =⎡⎢⎢⎢⎢⎢⎣

a11 ⋯ a1k⋮ ⋯ ⋮ak1 ⋯ akk

⎤⎥⎥⎥⎥⎥⎦(a) Show, using only the definition of a positive definite matrix (that is, without using

any characterization result of positive definite matrices)thataii > 0 for eachi ∈ {1, . . . ,n}.(b) Show, using only the definition of a positive definite matrix (that is, without using

any characterization result of positive definite matrices)that A(k) is a positive definitematrix for eachk ∈ {1, . . . ,n}.

(c) Show that the determinant ofA(k) is non-zero for eachk ∈ {1, ...,n}.(d) Show that the determinant ofA(k) is positive for eachk ∈ {1, ...,n}.


Solution.

(a) SinceA is positive definite,aii = (ei)′Aei > 0 for all i = 1, . . . ,n.

(b) For anyz∈Rk, we can write

z′A(k)z= [ z1 ⋯ zk ]⎡⎢⎢⎢⎢⎢⎣

a11 ⋯ a1k⋮ ⋱ ⋮ak1 ⋯ akk

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣

z1⋮zk

⎤⎥⎥⎥⎥⎥⎦= [ z1 ⋯ zk ]

⎡⎢⎢⎢⎢⎢⎣a11z1+⋅ ⋅ ⋅+a1kzk⋮ak1z1+⋅ ⋅ ⋅+akkzk

⎤⎥⎥⎥⎥⎥⎦= z1(a11z1+⋅ ⋅ ⋅+a1kzk)+⋅ ⋅ ⋅+zk(ak1z1+⋅ ⋅ ⋅+akkzk)=

k∑i=1

zi(z1ai1+⋅ ⋅ ⋅+zkaik)=

k∑i=1

zi⎛⎝

k∑j=1

zjai j⎞⎠

=k∑

i=1

k∑j=1

zizjai j

Let z∈Rk be such thatz≠ 0. Define the vectorzn ∈Rn by zni = zi if i ∈ {1, . . . ,k} and

zni = 0 if i ∈ {k+1, . . . ,n}. That is,

z= (z1, . . . ,zk)′zn = (z1, . . . ,zk,0, . . . ,0)′

Then we have

z′A(k)z= k∑i=1

k∑j=1

zizjai j

=n∑

i=1

n∑j=1

zizjai j (sinceznk+1 =⋯ = zn

n = 0)= (zn)′Azn

> 0 (sinceA is positive definite)Since our non-nullz∈ Rk is arbitrary, this shows thatA(k) is positive definite foreachk= 1, . . . ,n.


(c) [Note: It is fine to first show the stronger result requested in(d) and then state that theresult in (c) follows. The proof below establishes the result using only the definitionof positive definiteness, together with material discussedbefore Chapter 4.]

Suppose detA(k) = 0 for somek ∈ {1, . . . ,n}. ThenA(k) is singular, so the columnsof A(k) are linearly dependent. This means that there is some non-null z= (z1, . . . ,zk)′ ∈Rk such thatA(k)z=0. Premultiplying byz′, we have thatz′A(k)z=0 for somez≠0.But this is a contradiction ofA(k) being positive definite, which was established in(b). So detA(k) ≠ 0 for all k= 1, . . . ,n.

(d) It was shown in (b) thatA(k) is positive definite. From Section 4.6, we have theresult thatA(k) is positive definite if and only if every eigenvalueλ1, . . . ,λk of A(k)is positive. Then using the relation between characteristic values (eigenvalues) andthe determinant ofA(k) from Section 4.2, we have that detA(k) =∏k

i=1λi > 0.


ReferencesYou will find most of this material covered inElementary Matrix Algebraby F. Hohn

(Chapters 9, 10) andLinear Algebraby G. Hadley(Chapter 7). A more advanced treat-ment of the subject can be found inThe Theory of Matrices, vol. 1 byF.R. Gantmacher(Chapters 3, 10).

Part II

Real Analysis

69

Chapter 5

Basic Concepts of Real Analysis

5.1 Norm and Distance

We recall thatRn is the set of alln-vectorsx= (x1, ...,xn), where eachxi is a real numberfor i = 1, ...,n. The (Euclidean)normof a vectorx ∈Rn is denoted by∥x∥ and defined by

∥x∥ = [ n∑i=1

x2i ]

1/2

We have already noted some properties of the norm, wherex, y ∈Rnandλ ∈R:

(N.1) ∥x∥ ≥ 0; and∥x∥ = 0 iff x= 0.

(N.2) ∥λx∥ = ∣λ∣ ∥x∥(N.3) ∥x+y∥ ≤ ∥x∥+∥y∥

While (N.1) and (N.2) are easy to check, (N.3) is not. Start by establishing the follow-ing property:

(N.4) ∣xy∣ ≤ ∥x∥ ∥y∥The inequality (N.4) is known as theCauchy-Schwarz inequality. You can then use

(N.4) to prove (N.3).Using the norm, one can define the (Euclidean)distance functionor metric. Forx, y,

∈Rn, thedistancebetweenx andy, denoted byd(x,y), is

d(x,y) = ∥x−y∥The following properties ofd can be verified (whenx, y, z ∈Rn):

70

CHAPTER 5. BASIC CONCEPTS OF REAL ANALYSIS 71

(D.1) d(x,y) ≥ 0;d(x,y) = 0 if and only ifx= y

(D.2) d(x,y) = d(y,x)(D.3) d(x,z) ≤ d(x,y)+(d(y,z)

The property (D.3) is known as the “triangle inequality”; itcan be established using(N.3).

5.2 Open and Closed Sets

Open Ball:If x ∈Rn, andr is a positive real number, anopen ball(with centerx andradius r) in

Rn isB(x,r) = {x in R

n ∶ d(x,x) < r}Open Set:

A setS⊂Rn is open(in Rn) if for every x ∈ S, there is an open ball (with centerx andradiusr > 0) inRn which belongs toS.

It follows that an open ball is an open set. You can check that the setS= {(x1, x2) inR2 ∶ x1 > 0, x2 > 0 andx2

1+x22 < 1} is an open set inR2

.

In discussing the concept of an open set, it is important to specify thespacein whichwe are considering the set. For instance, the set{x ∈ R;0 < x < 1} is open inR; but theset{(x1, x2) ∈R2 ∶ 0< x1 < 1, x2 = 0} is not open inR2, although, graphically, the two sets“look the same”.Complement of a Set:

If S⊂ Rn, thecomplementof S (in Rn) is denoted by∼ S, and defined by∼ S= {x inRn ∶ x is not inS}Closed Set:

A setS⊂Rn is closedin Rn if the complement ofS in Rn is open inRn.You can check that the setS= {(x1, x2) in R2 ∶ x1 ≥ 0, x2 ≥ 0, andx2

1+x22 ≤ 1} is a closed

set inR2.There are many sets which areneitheropennor closed inRn. For example, the set

S= {(x1, x2) in R2 ∶ x1 ≥ 0, x2 ≥ 0, andx21+x2

2 < 1} is neither open nor closed inR2.If A⊂Rn andx ∈Rn, then one of three possibilities must hold:(1) There is an open ballB(x,r) such thatB(x,r) ⊂A.(2) There is an open ballB(x,r) such thatB(x,r) ⊂∼A.(3) If B(x,r) is any open ball, thenB(x,r) contains points of bothA and∼A.


Those pointsx ∈Rn which satisfy (1) constitute theinterior of A; those satisfying (2)theexterior of A; those satisfying (3) theboundaryof A. Points in these sets are called,respectively, interior, exterior and boundary points (with respect to the setA).Neighborhood:

If x ∈Rn, any set which contains an open set containingx is called aneighborhoodofx, and is denoted byN(x).

Thus, an open ball inRn with centerx and radiusr > 0 is a neighborhood ofx.Empty Set:

A setφ ⊂Rn which contains no elements ofRn is called theempty set.You can show thatRn andφ are the only two sets which are both open and closed in

Rn.

5.3 Convergent Sequences

Let x1, x2

, x3, ..... be a sequence of vectors inRn. A vectorx in Rn is called alimit of the

sequencex1, x2

, x3, ....if given any real numberε > 0, there is a positive integerN such that

d(xs, x) < ε whenevers>N. If the sequencex1

, x2, x3

, .... has a limit, we call the sequenceconvergent. If x is a limit of the sequence we say thatthe sequence convergesto x.

For example, the sequence of numbers 1,

12,

13,

14, .... is convergent, with limit equal to

zero.You can check that if a sequencex1

, x2, x3

, ...... is convergent, it has auniquelimit; soit makes sense to speak ofthe limit of a convergent sequence.

An important result on convergent sequences is that “weak inequalities are preservedin the limit”.

Proposition 2. Suppose{xs}∞1 is a convergent sequence of points inRn with limit x ∈Rn,and let a∈Rn. If xs≤ a for all s, then x≤ a.

Remark: You should be able to prove, by using Proposition 1, that ifb ∈Rn, and{ys}∞1is convergent of points inRn with limit y ∈Rn, andys≥ b for all s, theny≥ b.

Using the notion of convergent sequence, we can obtain a more“usable” characteriza-tion of closed sets.

Theorem 11. Let S⊂Rn. Then S is closed inRn iff whenever x1, x2, ..... is a sequence of

points of S that is convergent inRn, we have

limn→∞xn ∈S.


Proposition 2 combined with Theorem 11 gives us a convenientway of checking thatthe typical constraint sets which arise in optimization problems are closed. For example,you can check that the setS= {(x1, x2) ∈R2 ∶ 0≤ x1 ≤ 1, 0≤ x2 ≤ 2} is a closed set by usingProposition 2 and Theorem 11.

5.4 Compact Sets

Bounded Set:A setS⊂Rn is boundedif it is contained in some open ball inRn.For example, the setS= {(x1, x2) ∈R2+ ∶ x1 ≥ 0, x2 ≥ 0 andx1+x2 ≤ 1} is a bounded set,

because it is contained in the open ball inR2 with center 0 and radius 2.A setS⊂Rn is compactif it is both closed and bounded.

You can use the above definition to check that

(i) the setS1 = {(x1, x2) ∈R2 ∶ 0≤ x1 ≤ 1 and 0≤ x2 ≤ 2} is a compact set;

(ii) the setS2 = {(x1, x2) ∈R2 ∶ 0≤ x1 < 1 and 0≤ x2 < 1} is not a compact set, because itis not a closed set;

(iii) the setS3 = {(x1, x2) ∈R2 ∶ 0≤ x1 ≤ 1 and 0≤ x2) is not a compact set, because it isnot a bounded set.

5.5 Continuous Functions

Functions:Let A⊂Rn. A function, f , fromA toRm (written f ∶A→Rm) is a rule which associates

with each point inA a unique point inRm. In this caseA is called thedomainof f . Wedefine f (A) = {y ∈ Rm ∶ y = f (x) for somex ∈ A}. For example, iff (x) = x2 for all x ∈ R,thenA ≡ R, and f (A) = R+. In the special case wherem= 1, f is called areal valuedfunction.

If f ∶ A→ Rm is a function, we can definef 1(x) as the first component of the vectorf (x) for eachx ∈ A. Then f 1 is a function fromA to R. Similarly, f 2

, ..., f m can bedefined. These functionsf 1

, ...., f m are called thecomponent functionsof f .Conversely, ifg1

, ...,gm aremreal valued functions onA, we can defineg(x)= (g1(x), ...,gm(x))for eachx ∈A. Theng is a function fromA toRm.

Limit of a Function :


If f ∶A→R, anda ∈A, thenlimx→a

f (x) = b

means that given anyε > 0, there is a numberδ > 0, such that ifx ∈ A, and 0< d(x,a) < δ,then∣ f (x)−b∣ < ε.

For example, iff ∶R→R is defined byf (x) = x2−3, then limx→2

f (x) = 1.

Continuity of a Function :A function f ∶ A→R is calledcontinuous at a∈ A, if lim

x→af (x) = f (a). The functionf

is continuous(onA) if it is continuous at eachx ∈A.For example, iff ∶ R+ → R+ is defined byf (x) = x+1, then f is continuous at 0. If

g ∶R+→R+ is defined byg(x) = x+1 for x≠ 0, andg(0) = 0, theng is not continuous at 0.

5.6 Existence of Solutions to Constrained OptimizationProblems

The most important result to decide whether or not a constrained optimization problemhas a solution is known as Weierstrass theorem, and can be stated as follows:

Theorem 12. (Weierstrass) Suppose A is a non-empty closed and bounded subset ofRn.If f ∶ A→R is continuous on A, then there exist x1

, x2 in A such that f(x) ≤ f (x1) for allx ∈A and f(x) ≥ f (x2) for all x ∈A.


5.7 Appendix I: Closed Sets

Result:Let f ∶Rm

+ →R be a continuous function onRm+ . Then the set:

C= {x ∈Rm+ ∶ f (x) ≥ 0} (1)

is closed inRm.

Proof:Take an arbitrary convergent sequence{xn} of points inC, with limit z∈Rm. That is,

x1,x2

,x3, .... belong toC, and:

limn→∞xn = z (2)

We have to show thatz∈C.Sincexn ∈C for eachn = 1,2,3, ..., we havexn ≥ 0 for eachn = 1,2,3,⋯. Since (2)

holds, and weak inequalities are preserved in the limit, we havez≥ 0; that is,z∈Rm+ .

Sincef is continuous onRm+ , given anyε>0, there isδ>0, such that wheneverd(x,z)<

δ, we have: ∣ f (x)− f (z)∣ < ε (3)

Using thisδ > 0, we can find a positive integerN, such that whenevern > N, we haved(xn

,z) < δ, since (2) holds. Thus, using (3), for alln>N, we must have∣ f (xn)− f (z)∣ < ε.This implies that the sequence{ f (xn)} is convergent with limitf (z).

Sincexn ∈C for eachn = 1,2,3, ..., we have f (xn) ≥ 0 for eachn = 1,2,3,⋯. Since{ f (xn)} is convergent with limitf (z), and weak inequalities are preserved in the limit, wehave f (z) ≥ 0.

We have now shown thatz∈Rm+ , and f (z) ≥ 0, soz∈C.

Exercise 1:Let p ∈Rm

++, w> 0 and consider the budget set:

C= {x ∈Rm+ ∶ px≤w}

Show thatC is closed inRm, by defining f (x) =w− px for all x ∈Rm

+ , checking the conti-nuity of f onRm

+ , and then using the above result.

Exercise 2:Let u ∶Rm

+ →R be a continuous function onRm+ , and letv be a real number in the range

of the functionu. Consider the upper contour set:

C= {x ∈Rm+ ∶ u(x) ≥ v}


Show thatC is a closed set inRm, by defining f (x) = u(x)−v, checking the continuity off onRm

+ , and then using the above result.

Exercise 3:Let g j be a continuous function fromRm

+ toR, for eachj ∈ {1, ...,k}. Consider the set:

D = {x ∈Rm+ ∶ g j(x) ≥ 0 for eachj ∈ {1, ...,k}}

Show, by generalizing the argument used in establishing theabove result, thatD is a closedset inRm.

Sets likeD represent many of the constraint sets encountered in modernoptimizationtheory.


5.8 Appendix II: Continuity of Functions of Several Vari-ables

The sketch of continuity of the function:

f (x1,x2) = x1x2 for all (x1,x2) ∈R2

contains a more general idea, which is quite useful in checking continuity of some com-monly encountered functions of several variables. This general idea is developed below.

(a) LetA be a subset ofRn, and let f andg be functions fromA toR. Defineh ∶A→R

by:h(x) = f (x)g(x) for all x ∈A

The general result is the following: Iff andg are continuous functions onA, thenh is acontinuous function onA.

To prove this result, letz be an arbitrary point inA, and letε > 0 be given. We have toshow that wheneverx ∈A andd(x,z) < δ, we have:

∣h(x)−h(z)∣ < ε

Define:

ε′ =min{1, ε1+ ∣ f (z)∣+ ∣g(z)∣} (1)

Then 0< ε′ ≤ 1. Since f andg are continuous atz, given theε′ > 0, there existδ1 > 0 andδ2 > 0 such that wheneverx ∈A andd(x,z) < δ1, we have:

∣ f (x)− f (z)∣ < ε′ (2)

and wheneverx ∈A andd(x,z) < δ2, we have:

∣g(x)−g(z)∣ < ε′ (3)

Define:δ =min{δ1,δ2} (4)

Then, clearlyδ > 0.We can write forx ∈A,

∣h(x)−h(z)∣ = ∣ f (x)g(x)− f (z)g(z)∣= ∣ f (x)g(x)− f (x)g(z)+ f (x)g(z)− f (z)g(z)∣≤ ∣ f (x)(g(x)−g(z))∣+ ∣( f (x)− f (z))g(z)∣= ∣ f (x)∣∣(g(x)−g(z))∣+ ∣( f (x)− f (z))∣∣g(z)∣

(5)


Sincex ∈ A andd(x,z) < δ =min{δ1,δ2}, we know that both (2) and (3) must hold.Using this in (5), we obtain:

∣h(x)−h(z)∣ ≤ ∣ f (x)∣∣(g(x)−g(z))∣+ ∣( f (x)− f (z))∣∣g(z)∣< [1+ ∣ f (z)∣]ε′+ε′∣g(z)∣ ≤ ε

(6)

the inequalities on the last line of (6) following from (1). This establishes thath is contin-uous atz.

Remark:Note that, givenzandε, the appropriate definitions ofε′ and ofδ are suggested by (5).

So, (5) is really the first step in the proof, although it appears later in the formal proof,compared to the appearance ofε′ andδ.

(b) Let f ∶R2→R be defined by:

f (x1,x2) = x1 (7)

Then f is continuous onR2, since given anyz∈ R2, andε > 0, we can chooseδ = ε, and

note that wheneverx ∈R2 andd(x,z) < δ = ε, we must have∣x1−z1∣ < ε, and so∣ f (x1,x2)−f (z1,z2)∣ < ε.

(c) Letg ∶R2→R be defined by:

g(x1,x2) = x2 (8)

Theng is continuous onR2, since given anyz∈ R2, andε > 0, we can chooseδ = ε, andnote that wheneverx ∈R2 andd(x,z) < δ = ε, we must have∣x2−z2∣ < ε, and so∣g(x1,x2)−g(z1,z2)∣ < ε.

(d) Defineh ∶R2→R by:

h(x1,x2) = x1x2 for all (x1,x2) ∈R2

Then, we have:h(x1,x2) = f (x1,x2)g(x1,x2) for all (x1,x2) ∈R2

where f andg are defined by (7) and (8) respectively. Then, using the results in (a),(b) and(c), h is continuous onR2.


5.9 Appendix III: On a Variation of Weierstrass Theorem

Theorem:Let A be a non-empty subset ofRn

, and let f be a continuous function fromA to R.

SupposeB is a non-empty, closed and bounded set inRn, andz is an element ofB, such

thatB is a subset ofA, and:

f (x) ≤ f (z) f or all x ∈A∼B (1)

Then there is ¯x ∈A such that:

f (x) ≤ f (x) f or all x ∈A (2)

Proof:SinceB is a subset ofA and f is a continuous function fromA to R, f is continuous

on B. SinceB is a non-empty, closed and bounded set inRn, we can apply Weierstrass

theorem to infer that there is ¯x ∈B such that:

f (x) ≤ f (x) f or all x ∈B (3)

Let x be an arbitrary element ofA. There are two cases to consider: (i)x ∈B, (ii) x ∉B.In case (i), we have:

f (x) ≤ f (x) (4)

by (3).In case (ii), we havex ∈A∼B, and so:

f (x) ≤ f (z) (5)

by (1). Also, sincez∈B, we have:f (z) ≤ f (x) (6)

by (3). Combining (5) and (6), we obtain:

f (x) ≤ f (x) (7)

Thus, we have shown that in either case,f (x) ≤ f (x) must hold. Since ¯x ∈ B ⊂ A, weknow thatx ∈A, and this establishes (2).//

Remark:Note that in the statement of the theorem, the setA is assumed to be neither closed

nor bounded. However, the theorem ensures that there is a solution to the constrainedmaximization problem:

Maximize f(x)sub ject to x∈A

}(P)


This variation of Weierstrass theorem is often useful in applications.It is important to realize that the constrained maximization problem you will be given

will be in the form of(P); that is, you will be givenA and f . You arenot given B andz∈B. Thus, in order to apply the theorem to ensure that there exists a solution to(P), youhave to define an appropriate setB and an elementz∈ B with the properties stated in thetheorem.

Example:Let A=R+ and let f be a continuous function onR+, satisfying f (0) = 0 and f (x) ≤ 0

for all x> 1. Show that there is a solution to problem(P) by applying the above theorem.How would you chooseB andz∈B to apply the above theorem?

Converse of “Extension of Weierstrass Theorem ”Following example establishes the necessity of the Extension of Weierstrass Theorem.Example:Let A be a non-empty subset ofRn, and let f be a continuous function fromA to R.

Suppose there exists ¯x ∈A such that:

f (x) ≥ f (x) for all x ∈A. (8)

Then there exists a non-empty, closed and bounded set,B, in Rn, with B⊂ A, and there isz∈B, such that

f (z) ≥ f (x) for all x ∈A satisfying x ∉B (9)

Solution.

Note, intuitively, even though a closed ball may seem to be a candiadte for setB, wedo not know much about the setA, except that it is non-empty and has a global maximum.Thus it could be possible that the setA may not contain a closed ball at all. In the extremecase,A could be a singleton containing ¯x. Then our only choice would be to letB≡ {x},z=x. We will verify now that this choice is suitable in the general case also for any givenA.

Let B ≡ {x},z= x. SetB is bounded since any open ball with center ¯x and a positiveradius contains it.B is also closed since if we pick a convergent sequence fromB, all thepoints in this sequence must be ¯x, and terefore convergent to ¯x. SetB⊂A and f (z) ≥ f (x)for all x ∈A, x ∉B sincex is a global maximum off onA.

Thus we have proved existence ofB andz by construction.



Problem 21(Open Sets).

(a) LetSandT be open sets inRn. Show thatS∩T is also an open set inRn

.

(b) LetA be the set defined by:

A= {(x1,x2) ∈R2 ∶ x1 > 0,x2 > 0,x1x2 > 1}ExpressA as the intersection of two sets, and use (a) to show thatA is open inR2.

(c) LetB be the set defined by:

B= {(x1,x2) ∈R2 ∶ x1 ≥ 0,x2 ≥ 0,x1x2 > 1}Is B open inR2? Explain.

Solution.

(a) Consider an arbitrary ¯x ∈ S∩T. Now, since ¯x ∈ S andS is open inRn, there existsrS> 0 such thatB(x,rS) ⊂S. And since ¯x ∈ T andT is open inRn, there existsrT > 0such thatB(x,rT) ⊂ T. Take r =min{rS,rT} > 0. ThenB(x,r) ⊂ B(x,rS) ⊂ S andB(x,r) ⊂B(x,rT) ⊂ T. ThereforeB(x,r) ⊂S∩T, soS∩T is open inRn.

(b) We can writeA=A1∩A2 where

A1 = {(x1,x2) ∈R2 ∣ x1 > 0,x2 > 0}A2 = {(x1,x2) ∈R2 ∣ x1x2 > 1}

Note thatA2 contains some points wherex1 < 0 andx2 < 0. We want to show thatbothA1 andA2 are open inR2, which will show by part (a) thatA is open inR2.

Claim: A1 is open inR2.

Proof: Letx ∈A1 and taker =min{x1, x2} > 0. We want to showB(x,r) ⊂A1. This isthe same as showing that for anyx∉A1, we havex∉B(x,r). So, consider somex∉A1

and assume without loss of generality thatx1 ≤ 0. Then we have thatx1 ≤ 0< r ≤ x1,so x1− x1 ≥ r. That means thatd(x, x) =√(x1− x1)2+(x2− x2)2 ≥ r, so we havex ∉B(x,r), which is what we wanted to show. ThereforeA1 is open.

Claim: A2 is open inR2.


To show thatA2 is open inR2, it is easiest to show that∼A2 is closed inR2. First,define the functionf (x) = 1−x1x2 for all x ∈R2 and note thatf is continuous onR2.Then we can write∼A2 as

∼A2 = {(x1,x2) ∈R2 ∣ x1x2 ≤ 1} = {(x1,x2) ∈R2 ∣ f (x) ≥ 0}By Prof. Mitra’s result on closed sets, then,∼A2 is closed inR2. ThereforeA2 isopen inR2.

(c) We can write

B=A∪{(x1,x2) ∈R2 ∣ x1 = 0,x2 > 0,x1x2 > 1}∪{(x1,x2) ∈R2 ∣ x1 > 0,x2 = 0,x1x2 > 1}∪{(x1,x2) ∈R2 ∣ x1 = 0,x2 = 0,x1x2 > 1}

=A∪∅∪∅∪∅=A

SinceA is open inR2 andB=A, B is open inR2.

Problem 22(Closed Sets).

(a) LetSandT be open sets inRn. Show thatS∪T is also an open set inRn

.

(b) LetA be the set defined by:

A= {(x1,x2) ∈R2 ∶ x1 ≥ 0,x2 ≥ 0,x1+x2 ≤ 1}and letB be the complement ofA in R2. ExpressB as the union of three sets and show thatB is open inR2.

(c) Show thatA is closed inR2.

Solution.

(a) Consider an arbitrary ¯x ∈S∪T. Without loss of generality, assume ¯x ∈S. Now, sincex ∈ S andS is open inRn, there existsr > 0 such thatB(x,r) ⊂ S. But S⊂ S∪T, soB(x,r) ⊂S∪T. ThereforeS∪T is open inRn.

(b) We can write∼A=B=B1∪B2∪B3 where

B1 = {(x1,x2) ∈R2 ∣ x1 < 0}B2 = {(x1,x2) ∈R2 ∣ x2 < 0}B3 = {(x1,x2) ∈R2 ∣ x1+x2 > 1}


We want to show that each ofB1, B2, andB3 are open inR2, which will show bypart (a) thatB is open inR2.

Claim: B1 is open inR2.

Proof: Let x ∈ B1 and taker = −x1 > 0. We want to show thatB(x,r) ⊂ B1. Thisis the same as showing that for anyx ∉ B1, we havex ∉ B(x,r). So, consider somex ∉ B1. Then we have thatx1 ≥ 0, sox1− x1 = x1+ r ≥ r. That means thatd(x, x) =√(x1− x1)2+(x2− x2)2≥ r, so we havex∉B(x,r), which is what we wanted to show.ThereforeB1 is open.


Proof: Follow the steps from the proof above for the setB2 instead ofB1.


To show thatB3 is open inR2, it is easiest to show that∼B3 is closed inR2. First,define the functionf (x) = 1−x1−x2 for all x ∈R2 and note thatf is continuous onR2. Then we can write∼B3 as

∼B3 = {(x1,x2) ∈R2 ∣ x1+x2 ≤ 1} = {(x1,x2) ∈R2 ∣ f (x) ≥ 0}By Prof. Mitra’s result on closed sets, then,∼B3 is closed inR2. ThereforeB3 isopen inR2.

(c) We know thatB= ∼A. SinceB is open, by the definition of closed setsA is closed inR2.

Problem 23(Continuity of Functions).

Let f ∶Rn→R be a continuous function onRn

, and letx be a vector inRn, satisfyingf (x) > 0. Show that there is a positive real numberr such that:

f (x) > 0 for all x ∈B(x,r)Solution.

Let ε = f (x) > 0. Now, since f is continuous onRn, we know that given thisε =f (x) > 0 and anyx ∈ Rn, there is someδ > 0 such that wheneverd(x, x) < δ, we have∣ f (x)− f (x)∣ < f (x). This implies thatf (x)− f (x) < f (x) and f (x)− f (x) > − f (x), whichwe can rearrange for 0< f (x) < 2 f (x). Now, taker = δ and note that wheneverx ∈Rn andx ∈B(x,r), we haved(x, x) < δ and thusf (x) > 0, which is what we wanted to show.

Problem 24(Extension of Weierstrass theorem).


Let a andb be positive real numbers, and letf be a function fromR+ to R+, definedby:

f (x) = ax+b[x/(1+x)] for all x≥ 0

Consider the following constrained maximization problem:

Maximize f(x)−xsub ject to x≥ 0

}(P)(a) If a< 1, show that there exists a solution to problem(P).(b) If a≥ 1, show that there is no solution to problem(P).

Solution.

Defineg ∶R+→R by

g(x) = f (x)−x= (a−1)x+ bx1+x

for all x≥ 0

Then problem(P) is to maximizeg(x) subject tox≥0. The functiong is continuous on thenonempty and closed constraint setC= [0,∞), but we cannot use the Weierstrass Theorembecause the constraint set is not bounded.

(a) Note thatg is continuously differentiable onx≥ 0. The first order condition is

a−1+ b(1+x)2 = 0 Ô⇒ (1+x)2 = b1−a

This suggests using the Extension of Weierstrass Theorem onthe bounded setB =[0, b1−a]. We need to show that wheneverc> 1, we haveg( b

1−a) ≥ g(c b1−a).

g( b1−a) = −b+ b2

1−a

1+ b1−a

= −b+ b2

1−a+b=

ab−b1−a+b

g(c b1−a) = −bc+ b2c

1−a

1+ bc1−a

= −bc+ b2c1−a+bc

=abc−bc−b2c2+b2c

1−a+bc

=(ab−b)(c+1−1)−b2c(c−1)

1−a+bc=(ab−b)+(c−1)(ab−b−b2c)

1−a+bc

=ab−b−(c−1)b(bc+1−a)

1−a+bc

Wheneverc> 1, the numerator ing(c b1−a) is smaller and its denominator is larger,

so we haveg( b1−a) ≥ g(c b

1−a). Therefore we can apply the Extension of WeierstrassTheorem on the nonempty, closed, and bounded setB = [0, b

1−a] to conclude thatproblem(P) has a solution.


(b) For allx≥ 0, we have that

g′(x) = a−1+ b(1+x)2 > 0

sincea ≥ 1 andb > 0. Thereforeg is strictly increasing onx ≥ 0, so problem(P)cannot have a solution. Formally, we can suppose for the sakeof contradiction thatsome ¯x ≥ 0 is a solution to(P). Then definex′ = x+1 and note that becauseg isstrictly increasing onx≥ 0, we haveg(x′) > g(x). But this contradicts ¯x solving(P),so it must be that problem(P) has no solution.

Problem 25(Extension of Weierstrass Theorem).

Let p andq be arbitrary positive numbers, and letf ∶ R2+ → R be a continuous func-tion on R2+. Suppose there is(x1, x2) ∈ R2+ which satisfiesf (x1, x2) = 1. Consider theconstrained minimization problem:

Minimize px1+qx2

sub ject to f(x1,x2) ≥ 1and (x1,x2) ∈R2+

⎫⎪⎪⎪⎬⎪⎪⎪⎭(Q)Show that there is a solution to problem(Q).Solution.

First, defineg ∶R2+ →R by g(x1,x2) = f (x1,x2)−1 for all x ∈R2+. Since f is continu-ous onR2+, g is also continuous onR2+. Then by Prof. Mitra’s result on closed sets, theconstraint set

C= {(x1,x2) ∈R2 ∣ x1 ≥ 0,x2 ≥ 0,g(x1,x2) ≥ 0}is closed inR2. The objective function is continuous onC andC is nonempty because(x1, x2) ∈C. But, depending onf , C is not necessarily bounded. To use the Extension ofWeierstrass Theorem, define the set

B= {(x1,x2) ∈R2 ∣ x1 ≥ 0,x2 ≥ 0, f (x1,x2) ≥ 1, px1+qx2 ≤ px1+qx2}We could show thatB is closed by showing thatB is the intersection of two closed sets.Also, B is nonempty since(x1, x2) ∈B andB is bounded because it is contained in the open

ball B(0,r) wherer =max{ px1+qx2p ,

px1+qx2q }+1. Finally, for anyx′ ∈C,x′ ∉ B, we have

px′1+qx′2 > px1+qx2. Since(Q) is a minimization problem, it cannot be solved by anyx′ ∈C,x′ ∉ B. UsingB, then, the Extension of Weierstrass Theorem guarantees a solutionto (Q).


References:This material is standard in many texts on Real Analysis. You might consultPrinciples

of Mathematical Analysisby W. Rudin(Chapters 2, 3, 4) orIntroduction to Analysisby M.Rosenlicht(Chapters 3, 4). Some of the material is also covered inCalculus on Manifoldsby M. Spivak(Chapter 1).

Chapter 6

Differential Calculus

6.1 Partial Derivatives

Let A be an open set inRn, and letx ∈A. If f ∶A→R, the limit

limh→0

f (x1, ...,xi +h, ...,xn)− f (x1, ...,xn)h

if it exists, is called theith (first-order)partial derivativeof f at x, and is denoted byDi f (x), wherei = 1, ...,n. This means, of course, that we can compute partial derivativesjust like ordinary derivatives of a function ofonevariable. That is, iff (x1, ...,xn) is givenby some formula involving(x1, ...,xn), then we findDi f (x) by differentiating the functionwhose value atxi is given by the formula when allx j (for j ≠ i) are thought of as constants.

For example, if f ∶ R2→ R is given by f (x1, x2) = x3

1+3x32+2x1x2, then D1 f (x) =

3x21+2x2, D2 f (x) = 9x2

2+2x1.The ith partial derivative off at x is also, alternatively, denoted byfi(x). When the

ith partial derivative off at x exists for alli = 1, ...,n, we can define then-vector of thesepartial derivatives ∇ f (x) ≡ [D1 f (x), ...,Dn f (x)]This vector is called thegradient vectorof f at x. In the example described above, thegradient vector is

∇ f (x) ≡ [3x21+2x2, 9x2

2+2x1] for all x ∈R2

When f ∶ A→ R has (first-order) partial derivatives at eachx ∈ A, we say thatf has(first-order)partial derivatives on A.

Second-Order Partial Derivatives and the Hessian Matrix

87

CHAPTER 6. DIFFERENTIAL CALCULUS 88

When f ∶A→R has (first-order) partial derivatives onA, these first-order partial deriva-tives are themselves functions fromA to R. If these (first-order) partial derivatives arecontinuous onA, then we say thatf is continuously differentiableonA. If these functionshave (first-order) partial derivatives onA, thesepartial derivatives are called thesecond-order partial derivativesof f onA.

To elaborate, ifDi f (x) exists for allx ∈ A, we can define the functionDi f ∶ A→ R.If this function has (first-order) partial derivatives onA, then the jth (first-order) partialderivative ofDi f at x [that is,D j(Di f (x))] is a second-order partial derivative off at x,and is denoted byDi j f (x). [Herei = 1, ...,n and j = 1, ...,n].

In the example described above,D11 f (x) = 6x1; D12 f (x) = 2 = D21 f (x); D22 f (x) =18x2. We note in this example that the “cross partials”D12 f (x) andD21 f (x) are equal.This is not a coincidence; it is a more general phenomenon as noted in the following result,known as “Young’s theorem”.

Theorem 13. (Young) Suppose A is an open set inRn, and f has first and second-orderpartial derivatives on A. If Di j f and Dji f are continuous on A, then Di j f (x) =D ji f (x)for all x ∈A

When all of the hypotheses of Theorem 1 hold, we will say thatf is twice continuouslydifferentiableonA; this will be the typical situation in many applications.

When the first and second-order partial derivatives off ∶ A→ R exist onA, then×nmatrix of second-order partial derivatives off described below:

H f (x) =⎡⎢⎢⎢⎢⎢⎣

D11 f (x) D12 f (x)...D1n f (x)- - - - - - - - - - - - - - - - -Dn1 f (x) Dn2 f (x)...Dnn f (x)

⎤⎥⎥⎥⎥⎥⎦is called theHessian matrixof f at x ∈ A, and is denoted byH f (x). When f is twicecontinuously differentible onA, the Hessian matrix off is symmetricat allx ∈A.

In the example described above

H f (x) = [ 6x1 22 18x2

]is the Hessian matrix off for all (x1, x2) ∈R2.

6.2 Composite Functions and the Chain Rule

Let g ∶ A→Rm be a function with component functionsgi ∶ A→R(i = 1, ...,m) which aredefined on an open set A⊂ Rn. Let f ∶ B→ R be a function defined on an open set


B⊂Rm which contains the setg(A). Then, we can defineF ∶ A→R by F(x) ≡ f [g(x)] ≡f [g1(x), ...,gm(x)] for eachx ∈ A. This function is known as acomposite function[of fandg].

The “Chain Rule” of differentiation provides us with a formulafor finding the partialderivatives of a composite function,F , in terms of the partial derivatives of the individualfunctions, f andg.

Theorem 14. (Chain Rule) Let g∶A→Rm be a function with component functions gi ∶A→R(i = 1, ...,m) which are continuously differentiable on an open set A⊂Rn. Let f ∶B→R

be a continuously differentiable function on an open set B⊂ Rm which contains the setg(A). If F ∶A→R is defined by F(x) = f [g(x)] on A, and a∈A, then F is differentiable ata and we have for i= 1, ...,n,

DiF(a) = m∑j=1

D j f (g1(a), ...,gm(a))Dig j(a)Examples:(i) Herem= 2, n= 1. Letg1(x) = x2 onR, andg2(x) = 1+x onR; let f (y1,y2) = y1+y2

2 onR2. ThenF(x)= f [g(x)]= f [g1(x), g2(x)]=g1(x)+[g2(x)]2= x2+(1+x)2 is a compositefunction onR. If a ∈R,

F ′(a) = D1F(a) =D1 f (g1(a), g2(a)) ⋅D1g1(a)+D2 f (g1(a), g2(a)) ⋅D1g2(a)= 2a+2g2(a) = 2a+2(1+a)

(ii) Here m= 1, n = 2. Let g1(x) = g1(x1, x2) = x21+ x2 on R2; f (y) = 4y on R. Then

F(x) = F(x1, x2) = f [g1(x1, x2)] = 4[x21+x2]. Then ifa ∈R2,

D1F(a) = D1 f [g1(a1, a2)]D1g1(a1, a2)D2F(a) = D1 f [g1(a1, a2)]D2g1(a1, a2)

Thus,D1F(a) = 4(2a1); D2F(a) = 4(1).6.3 Homogeneous Functions and Euler’s Theorem

A functionF ∶Rn+→R is homogeneous of degree ronRn

+ if for all x in Rn+, and allt > 0,

f (tx) = tr f (x)


Considerf ∶ R2+ → R given by f (x1, x2) = xa1xb

2 wherea > 0 andb > 0. Then if t > 0, wehavef (tx1, tx2)= (tx1)a(tx2)b= ta+bxa

1xb2= ta+b f (x1, x2). So, f is homogeneous of degree(a+b).

We can calculate the partial derivatives off on R2++. Thus,D1 f (x1,x2) = axa−11 xb

2;D2 f (x1,x2)=bxa

1,xb−12 . Now, if t >0, thenD1 f (tx1,tx2)=a(tx1)a−1(tx2)b = ta+b−1axa−1

1 xb2=

ta+b−1D1 f (x1,x2). SoD1 f is homogeneous of degree(a+b−1). Similarly, one can checkthatD2 f is homogeneous of degree(a+b−1). More generally, whenever a function,f , ishomogeneous of degreer, its partial derivatives are homogeneous of degree(r −1) (undersuitable differentiability assumptions), and this is demonstrated in Theorem 15 below.

We can verify thatx1D1 f (x1, x2)+x2D2(x1, x2) = axa1xb

2+bxa1xb

2 = (a+b)xa1xb

2 = (a+b) f (x1, x2). More generally, when a function,f , is homogeneous of degreer, then (undersuitable differentiability assumptions)x∇ f (x) = r f (x), a result known as Euler’s theorem,which we prove below in Theorem 16.

Theorem 15. Suppose f∶ Rn+ → R is homogeneous of degree r onRn

+, and continuouslydifferentiable onRn

++. Then for each i= 1, ...,n,Di f is homogeneous of degree (r−1) onRn++.

To prove this lett > 0 be given and define the functiong1, ...,gn fromRn

++ toR by

gi(x) = txi i = 1, ...,n

and the composite functionF from Rn++ to R by

F(x) = f [g1(x), ...,gn(x)] = f (tx1, ....,txn)Then applying the Chain Rule, we have for eachi = 1, ...,n

DiF(x) =Di f (tx1, ...,txn) ⋅ t (6.1)

But sincef is homogeneous of degreer, we have

F(x) = tr f (x1, ...,xn)So,

DiF(x) = trDi f (x1, ...,xn) (6.2)

Using (6.1) and (6.2), we get

Di f (tx1, ...,txn) = tr−1Di f (x1, ...,xn) (6.3)

sincet > 0.


Theorem 16. (Euler’s Theorem) Suppose f∶Rn+→R is homogeneous of degree r onRn

+and continuously differentiable onRn

++. Then

x∇ f (x) = r f (x) f orall x ∈Rn++

To prove this, let(x1, ...,xn) be given inRn++, and define the functionsg1

, ...,gn fromR++ toR+ by

gi(t) = txi , i = 1, ...,n

and the composite functionF from R++ to R by

F(t) = f [g1(t), ...,gn(t)] = f (tx1, ...,txn)Then, applying the Chain Rule, we have

F ′(t) ≡D1F(t) = n∑i=1

Di f [g1(t), ...,gn(t)]xi (6.4)

But sincef is homogeneous of degreer, we have

F(t) = tr f (x1, ...,xn)and,

F ′(t) ≡D1F(t) = rt r−1 f (x1, ...,xn) (6.5)

Also, for i = 1, ....,n we have

Di f [g1(t), ...,gn(t)] =Di f [tx1, ...,txn] = tr−1Di f (x1, ...,xn) (6.6)

by using Theorem 15. Thus, combining (6.4), (6.5) and (6.6),

rt r−1 f (x1, ...,xn) = n∑i=1

tr−1Di f (x1, ...,xn)xi (6.7)

Cancelling the common termtr−1 > 0 on both sides of (6.7), we get

r f (x1, ...,xn) = n∑i=1

Di f (x1, ...,xn)xi

which is the desired result.


6.4 The Inverse and Implicit Function Theorems

Jacobians:SupposeA is an open set inRn, and f is a function fromA to Rn, with component

functions f 1, ..., f n. If a ∈ A, and the partial derivatives off 1

, ...., f n exist ata, then then×n matrix

D f (a) ≡ (D j f i(a))is defined as theJacobian matrixof f at a. The determinant of this matrix, denoted byJf (a), is defined as theJacobianof f ata.

For example, if f 1(x1, x2) = x21+2x1x2 on R2, and f 2(x1, x2) = x1+3x3

2 on R2, anda= (a1, a2) is in R2, then

D f (a) = [ 2a1+2a2 2a1

1 9a22]

is the Jacobian matrix ata, and the Jacobian ata is Jf (a) ≡ [18a32+18a1a2

2−2a1]. Notethat, typically, the Jacobian matrix isnot a symmetric matrix (unlike a Hessian matrix).

Inverse Functions:Let A be a set inRm

, and let f be a function fromA to Rn. Then f is one-to-oneon

A if wheneverx1, x2 ∈ A andx1 ≠ x2, we havef (x1) ≠ f (x2). If there is a functiong, from

f (A) to A, such thatg[ f (x)] = x for eachx∈A, theng is called theinverse functionof f onf (A).

If f ∶R→R is defined byf (x)=2x, then we note thatf is one-to-one onR; also we candefine the functiong ∶R→R by g(y)= (1/2)y, and note that it has the propertyg[ f (x)]= x;g is then theinverse functionof f onR. Furthermoreg′[ f (x)] = 1/ f ′(x) for all x ∈R.

More generally, letA be an open set inR, and f ∶A→R be continuously differentiableonA. Let a ∈A, and suppose thatf ′(a) ≠ 0. If f ′(a) > 0, then there is an open ballB(a,r)such thatf ′(x) > 0 for all x in B(a,r), and f is increasing onB(a,r). Thus, for everyy ∈ f [B(a,r)], there is aunique xin B(a,r) such thatf (x) = y. That is, there is a uniquefunction g ∶ f [B(a,r)]→ B(a,r) such thatg[ f (x)] = x for all x ∈ B(a,r). Thus,g is aninverse function off on f[B(a,r)}; we say thatg is the inverse off “locally” around thepoint f (a). [Notice that there is no guarantee that the inverse functionis defined on theentire setf (A)]. Similarly, if f ′(a) < 0, an inverse function could be defined “locally”around f (a). The important restriction to carry out the kind of analysis noted above isthat f ′(a) ≠ 0.

To illustrate this, considerf ∶ R→ R+ given by f (x) = x2. Consider the pointa = 0.

Clearly f is continuously differentiable onR, but f ′(a) = f ′(0) = 0. Now, we cannotdefine a unique inverse function off even “locally” aroundf (a). That is, choose any


open ballB(0,r), and consider any pointy ≠ 0 in the setf [B(0,r)]. There will betwovaluesx1, x2 in B(0,r), x1 ≠ x2, such thatf (x1) = y and f(x2) = y.

Of course,f ′(a) ≠ 0 is not a necessary condition to get a unique inverse function off . For example iff ∶ R→ R is defined byf (x) = x3, then we havef to be continuouslydifferentible onR, with f ′(0) = 0. However f is an increasing function, and clearly has aunique inverse functiong(y) = y1/3 onR, and hence locally aroundf (0).Theorem 17. (Inverse Function Theorem) Let A be an open set ofRn, and f ∶ A→Rn becontinuously differentiable on A. Suppose a∈ A and the Jacobian of f at a is non-zero.Then there is an open set X⊂A containing a, and an open set Y⊂Rn containing f(a), anda unique function g∶Y→X, such that:

(i) f (X) =Y;(ii) f is one-to-one on X;(iii) g(Y) =X, and g[ f (x)] = x for all x ∈X.

Further, g is continuously differentiable on Y.

Note that (i) and (iii) of Theorem 17 imply thatg is an inverse function off on f (X).There are two preliminary implications of the theorem that are worth noting.

First, g is one-to-one onY. To see this, lety1,y2 ∈Y, and suppose thatg(y1) = g(y2).

Denote the common vector byx; by (iii) of Theorem 17,x ∈ X. Using (i) of Theorem 17,we can findx1

,x2 ∈X such thatf (x1) = y1 and f (x2) = y2. By (iii) of Theorem 17,

x = g(y1) = g( f (x1)) = x1

x = g(y2) = g( f (x2)) = x2

Thus, we must havex1 = x2 = x, and consequentlyy1 = y2. This establishes thatg is one-to-

one onY.Second, we must have:

f (g(y)) = y f or all y ∈Y (1)

To see this, lety be an arbitrary vector inY. By (iii) of Theorem 17,g(y) ∈X, and by (i) ofTheorem 17,f (g(y)) ∈Y. Denoteg(y) by x and f (g(y)) by z. Then, by (iii) of Theorem17, we haveg( f (x)) = x. But, g( f (x)) = g( f (g(y))) = g(z), by definition ofx andz. Thus,g(z) = x. But, by definition ofx, we haveg(y) = x. Thus, using the fact thatg is one-to-oneonY, we must havey= z. This establishes (1).

The advantage of noting both the identity in (iii) of Theorem17 and the identity indisplay (1) is that, depending on the application, one mightuse either form to obtain thepartial derivatives of the inverse function,g, at f (a).

We illustrate how these partial derivatives can be obtainedby using the identity in (iii)of Theorem 17. We can define forx ∈ X, F1(x) = g1[ f (x)] as a composite function off


andg1. Using the Chain Rule we get:

DiF1(x) = n∑j=1

D jg1[ f (x)]Di f j(x) f or i = 1, ...,n

But sinceF1(x) = x1, we haveDiF1(x) = 1 for i = 1, while DiF1(x) = 0 for i ≠ 1. Wecan repeat these calculations withF2(x) = g2[ f (x)], and getDiF2(x) = 1 for i = 2, whileDiF2(x) = 0 for i ≠ 2. The results forF3(x), ...,Fn(x) should now be obvious. Thisinformation can then be written in familiar matrix multiplication form:

I =

⎡⎢⎢⎢⎢⎢⎣D1g1[ f (x)]....Dng1[ f (x)]−−−−−−−−−−−−D1gn[ f (x)]....Dngn[ f (x)]

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎢⎢⎣

D1 f 1(x)....Dn f 1(x)D1 f 2(x)....Dn f 2(x)- - - - - - - - - - - -D1 f n(x)....Dn f n(x)

⎤⎥⎥⎥⎥⎥⎥⎥⎦That is,I =Dg[ f (x)]D f (x). Thus,D f (a) is invertible, and we haveDg[ f (a)]= [D f (a)]−1.This yields, in turn,Jg[ f (a)] = 1/Jf (a), sinceJf (a) ≠ 0.Example:

Let f ∶R2→R2 given by f (x,y) = (y, x+y2) for (x,y) ∈R2. Let us consider the point(a,b) = (1,1). The Jacobian matrix off at (1,1) is

D f (1,1) = [ 0 11 2

]The Jacobian off at (1,1) is

Jf (1,1) = −1You can check thatf is continuously differentiable onR2. Thus, we can invoke theinverse function theorem and get an open setA containing(1,1), an open setB containingf (A), and a unique functiong ∶B→A, such thatg is continuously differentiable onB, andg[ f (x,y)] = (x,y) for all (x,y) in A. Sog1(y, x+y2) = x, andg2(y, x+y2) = y for all (x,y)in A.

Let (Z1, Z2) ∈ B. Then we can definey= Z1, x= Z2−Z21. Theng1(Z1, Z2) = g1(y, x+

y2) = x = Z2 −Z21. Thus g1(Z1, Z2) = Z2 −Z2

1 for (Z1, Z2) in B. Similarly we haveg2(Z1, Z2) = g2(y, x+ y2) = y = Z1 for (x1, x2) in B. Thus, in this case we can actuallycomputethe inverse function. We can calculate the Jacobian matrix of g at f (1,1)= (1,2) ∶

Dg[ f (1,1)] = [ −2 11 0

]


It can be checked easily that

Dg[ f (1,1)]D f (1,1) = I

by simply multiplying the two matrices.

Implicit Functions :Consider the functionf ∶ R2

→ R given by f (x,y) = x2+y2−1. If we choose(a,b)with f (a,b) = 0, anda≠ 1, a≠ −1, there are open intervalsX ⊂R containinga, andY ⊂Rcontainingb, such that ifx∈X, there is aunique y∈Y with f (x,y) =0. Thus, we can definea unique functiong ∶ X→Y such thatf (x,g(x)) = 0 for all x ∈ X. [If a> 0 andb> 0, theng(x) = (1−x2)1/2 on X.] Such a function is said to be definedimplicitly by the equationf (x,y) = 0. Note that ifa= 1, andb= 0, so thatf (a,b) = 0, wecannotfind such a uniquefunction,g.

The above example can be generalized considerably to obtaina very important result,which is known as the Implicit Function Theorem.

Theorem 18. (Implicit Function Theorem) Let A be an open set ofRn+m, and let f1, ..., f m

be continuously differentiable functions from A toR. Let(a;b) ∈A, with a∈Rn and b∈Rm,such that fi(a;b) = 0 for i = 1, ...,m. Suppose the m×m matrix Dn+ j f i(a;b)[i = 1, ...,mand j= 1, ...,m] has a non-zero determinant. Then there exists an open set X containinga and an open set Y containing b, and a unique function g∶X→Y, such that

(i) f (x,g(x)) = 0 for all x ∈X(ii) g(a) = b

Further, g is continuously differentiable on X.Example:

Consider the function,f ∶R2++→R, defined byf (x,y) = xαyβ−1 for x> 0, y> 0, whereα,β are positive constants. Note that at(a,b) = (1,1), we havef (a,b) = f (1,1) = 0. Wecan calculate the (one by one) matrixD2 f (a,b) =D2 f (1,1) = β, and this has a determinantequal toβ≠0. Sincef is continuously differentiable onR2++, so we can invoke the implicitfunction theorem, to obtain an open setA containinga, and an open setB containingb,and a unique functiong ∶A→B, such thatg(a) =b, and f (x,g(x)) =0 for all x∈A. Further,g is continuously differentiable onA. Thus, definingF ∶ A→R by F(x) = f (x,g(x)), wehave by the chain-rule

F ′(x) =D1 f (x,g(x))+D2 f (x,g(x))g′(x)But sinceF(x) = 0 for all x ∈A, we get

0=D1 f (x,g(x))+D2 f (x,g(x))g′(x)


Transposing terms, and evaluating the derivatives atx= a,

−g′(a) = D1 f (a,b)D2 f (a,b) sinceD2 f (a,b) ≠ 0

Thus, −g′(1) = [D1 f (1,1)/D2 f (1,1)] = (α/β).



Problem 26(Converse of Euler’s Theorem).

(a) Let f ∶Rn++ →R++ be a continuously differentiable function on its domain, which

satisfiesx∇ f (x) = f (x)

for all x ∈Rn++. Show thatf is homogeneous of degree one on its domain.

(b) Generalize the result in (a) to provide an appropriate converse of Euler’s theorem.

Solution.

The proofs of parts (a) and (b) are very similar, so let’s construct and prove the moregeneral result requested in part (b). Then part (a) will follow for the special case where thedegree of homogeneity is one.

Claim: Let f ∶ Rn++ → R++ be a continuously differentiable function onRn

++, and sup-pose thatx∇ f (x) = r f (x) for all x ∈Rn

++. Then f is homogeneous of degreer onRn++.

Proof: Let x be an arbitrary vector inRn++. We want to show that for allt > 0, we have

f (tx) = tr f (x). Given this ¯x ∈Rn++, define the functiong ∶R++→R++ by

g(t) = f (tx1, . . . ,txn) for all t > 0

Using the Chain Rule and the hypothesisx∇ f (x) = r f (x), we have that for allt > 0,

g′(t) = n∑i=1

Di f (tx1, . . . ,txn)xi = x∇ f (tx) = rt

f (tx) = rtg(t)

Rearranging, we have thattg′(t) = rg(t) for all t > 0

Now, considerg(t)tr . Differentiating with respect tot, we have that for allt > 0,

∂∂t(g(t)

tr ) = g′(t)tr −g(t)rt r−1

t2r =tr−1

t2r (tg′(t)− rg(t)) = 0

This implies that for allt > 0, g(t)tr = c for somec ∈ R. Evaluating att = 1, we have that

c= g(1) = f (x), so g(t)tr = f (x). Sinceg(t) = f (tx), we have thatf (tx) = tr f (x) for all t > 0,

which is what we wanted to show.


Problem 27(Homothetic Functions).

A functionF ∶Rn+→R+ is called ahomothetic functiononRn

+ if there exists a functionf ∶ Rn

+ → R+ which is homogeneous of degree one onRn+, and there exists a function

g ∶R+→R+ which is an increasing function onR+, such thatF(x) = g( f (x)) for all x∈Rn+.

(a) LetF ∶Rn+→R+ be a function which is homogeneous of degreer > 0 onRn

+. ShowthatF is a homothetic function onRn

+.(b) LetF ∶R2+→R+ be defined by:

F(x1,x2) = xα1x1−α

2

1+xα1x1−α

2

for all (x1,x2) ∈R2+

whereα ∈ (0,1) is a parameter. Verify thatF is a homothetic function onR2+.(c) Supposef ∶R2++ →R++ is a continuously differentiable function onR2++, which is

homogeneous of degree one, and which satisfiesD1 f (x) > 0 andD2 f (x) > 0 for all x inR2++. Supposeg ∶R++→R++ is a continuously differentiable function satisfyingg′(y) > 0for all y in R++. Let F ∶ R2++ → R++ be defined byF(x) = g( f (x)). If t is an arbitrarypositive real number, show that

[D1F(tx)/D2F(tx)] = [D1F(x)/D2F(x)]for all x in R2++.

Solution.

Note that part (a) shows that all homogeneous functions are also homothetic functions,while part (b) shows that a homothetic function is not necessarily homogeneous.

(a) Homogeneity of degreer of F requiresF(tx)= trF(x), which is equivalent tog(t f (x))=trg( f (x)) since f must be homogeneous of degree one. This suggests choosingg(x) = xr . That choice impliesF(x) = ( f (x))r , which suggests choosingf (x) =(F(x)) 1

r . Now we must formally use these choices to show thatF is homothetic.

Define f ∶Rn+→R+ andg ∶R+→R+ by

f (x) = (F(x)) 1r for all x ∈Rn

+, g(x) = xr for all x ∈R+

Note thatg is increasing onR+ because we are given thatr > 0. Also, f is homoge-neous of degree one onRn

+ because for anyx ∈Rn+ and anyt > 0, we have

f (tx) = (F(tx)) 1r = (trF(x)) 1

r = t(F(x)) 1r = t f (x)


Finally, verify that for allx ∈Rn+,

g( f (x)) = g((F(x)) 1r ) = ((F(x)) 1

r )r = F(x)By the given definition, then,F is homothetic onRn

+.

(b) Let f ∶R2+→R+ be defined by

f (x1,x2) = xα1x1−α

2 for all (x1,x2) ∈R2+

Now, f is homogeneous of degree one onR2+ because for any(x1,x2) ∈R2+ and anyt > 0, we have

f (tx1,tx2) = tαt1−αxα1x1−α

2 = t f (x1,x2)Now, letg ∶R+→R+ be defined by

g(x) = x1+x

for all x≥ 0

Note thatg is increasing onR+ because for allx≥ 0, we have

g′(x) = 1+x−x(1+x)2 = 1(1+x)2 > 0

Finally, verify that for all(x1,x2) ∈R2+,

g( f (x1,x2)) = g(xα1x1−α

2 ) = xα1x1−α

2

1+xα1x1−α

2

= F(x1,x2)By the given definition, then,F is homothetic onR2+.

(c) Since f is homogeneous of degree one onR2++, we have by Theorem 15, that thefirst-order partial derivatives off are homogeneous of degree zero onR2++. That is,for anyx ∈R2++ and anyt > 0, we have

D1 f (tx) =D1 f (x), D2 f (tx) =D2 f (x)Now, using this result after applying the Chain Rule toF , we have that for anyx ∈R2++ and anyt > 0,

D1F(tx) = g′( f (tx))D1 f (tx) = g′( f (tx))D1 f (x) (2.1)

D2F(tx) = g′( f (tx))D2 f (tx) = g′( f (tx))D2 f (x) (2.2)


Because we are given that the first order partial derivatives of F are positive, we candivide (2.1) by (2.2) to see that for anyx ∈R2++ and anyt > 0,

D1F(tx)D2F(tx) = g′( f (tx))D1 f (x)

g′( f (tx))D2 f (x) = D1 f (x)D2 f (x) (2.3)

Evaluating att = 1, we have that for anyx ∈R2++,

D1F(x)D2F(x) = D1 f (x)

D2 f (x) (2.4)

Combining (2.3) and (2.4), we have that for anyx ∈R2++ and anyt > 0,

D1F(tx)D2F(tx) = D1F(x)

D2F(x)Problem 28(Inverse Function Theorem).

Supposegi (for i = 1,2) are continuously differentiable functions fromR2++ to R++,satisfying for all(x1,x2) ∈ R2++ , D jgi(x1,x2) > 0 for i = 1,2 ; j = 1,2 . Define functionsai j (x1,x2) =D jgi(x1,x2) for (x1,x2) ∈R2++ . Let x0 = (x0

1,x02) ∈R2++ ; denotegi(x0) by y0

ifor i = 1,2 . Assume that

a11(x0)a22(x0) > a12(x0)a21(x0)Define the functionsf 1 and f 2 from R4++ to R++ as follows:

f 1(x1,x2,y1,y2) = y1−g1(x1,x2)f 2(x1,x2,y1,y2) = y2−g2(x1,x2)

(a) Use the implicit function theorem to show that there exists an open setU containingy0 = (y0

1,y02), and an open setV containingx0, and a unique functionh ∶U →V such that

for all y= (y1,y2) ∈U , g(h(y)) = y, andh(y0) = x0.(b) Note thath is continuously differentiable onU , and show thatD1h1(y0) > 0 , and

D1h2(y0) < 0.

Solution.

Note that in this problem,x1 andx2 are treated as variables andy1 andy2 are treatedas parameters. This is reversed from the presentation of theImplicit Function Theorem inclass.


(a) Note thaty0 ∈ R2++ becauseg1 andg2 are functions intoR++. We want to applythe Implicit Function Theorem tof 1 and f 2 at the point(x0

,y0) ∈R4++. We need tocheck some conditions before applying the theorem:

● Both f 1 and f 2 are defined on the open setR4++. We would like to use the factthat each off 1 and f 2 is the sum of two continuously differentiable functionsonR4++ to establish thatf 1 and f 2 are continuously differentiable onR4++, butg1 andg2 are defined onR2++. Therefore, define the following functions, eachfrom R4++ to R++:

g1(x1,x2,y1,y2) = g1(x1,x2) for all (x1,x2,y1,y2) ∈R4++

g2(x1,x2,y1,y2) = g2(x1,x2) for all (x1,x2,y1,y2) ∈R4++

g3(x1,x2,y1,y2) = y1 for all (x1,x2,y1,y2) ∈R4++

g4(x1,x2,y1,y2) = y2 for all (x1,x2,y1,y2) ∈R4++

Each ofg1, g2

, g3, andg4 is continuously differentiable onR4++. We can expresseach off 1 and f 2 as sums of these functions:

f 1(x1,x2,y1,y2) = g3(x1,x2,y1,y2)− g1(x1,x2,y1,y2)f 2(x1,x2,y1,y2) = g4(x1,x2,y1,y2)− g2(x1,x2,y1,y2)

Thereforef 1 and f 2 are continuously differentiable onR4++.

● We are given that at(x0,y0) ∈R4++,

f 1(x0,y0) = y0

1−g1(x0) = y01−y0

1 = 0

f 2(x0,y0) = y0

2−g2(x0) = y02−y0

2 = 0

● By the definition off 1 and f 2 we have that

(D j f i(x0,y0)) = [ D1 f 1(x0

,y0) D2 f 1(x0,y0)

D1 f 2(x0,y0) D2 f 2(x0

,y0) ]= [ −D1g1(x0) −D2g1(x0)−D1g2(x0) −D2g2(x0) ]= [ −a11(x0) −a12(x0)−a21(x0) −a22(x0) ]

Then det(D j f i(x0,y0)) = a11(x0)a22(x0)−a12(x0)a21(x0) ≠ 0 by the given in-

formation.


By the Implicit Function Theorem, then, there is an open setU ⊂R2++ containingy0,an open setV ⊂R2++ containingx0, and a unique functionh ∶U →V such that:

(1) f (h(y),y) = 0 for all y ∈U

(2) h(y0) = x0

Now, result (1) implies that for eachi = 1,2, we havef i(h(y),y) = yi −gi(h(y)) = 0for all y ∈U . This implies that for ally ∈U , we have

g1(h1(y),h2(y)) = y1 (3.1)

g2(h1(y),h2(y)) = y2 (3.2)

This is equivalent tog(h(y)) = y for all y ∈U , which we were asked to show.

(b) Given the conditions that were checked in part (a), we also have from the ImplicitFunction Theorem thath is continuously differentiable onU .

Using the Chain Rule to differentiate (3.1) with respect to each of y1 andy2, we havethat for ally ∈U ,

D1g1(h1(y),h2(y))D1h1(y)+D2g1(h1(y),h2(y))D1h2(y) = 1 (3.3)


Using the Chain Rule to differentiate (3.2) with respect to each of y1 andy2, we havethat for ally ∈U ,



Evaluating (3.3), (3.4), (3.5), and (3.6) aty0 and using the result from part (a) thath(y0) = x0, we have the four equations

D1g1(x0)D1h1(y0)+D2g1(x0)D1h2(y0) = 1

D1g1(x0)D2h1(y0)+D2g1(x0)D2h2(y0) = 0

D1g2(x0)D1h1(y0)+D2g2(x0)D1h2(y0) = 0

D1g2(x0)D2h1(y0)+D2g2(x0)D2h2(y0) = 1

This is a system of linear equations that we can express as

[ a11(x0) a12(x0)a21(x0) a22(x0) ][ D1h1(y0) D2h1(y0)

D1h2(y0) D2h2(y0) ] = [ 1 00 1

]


In particular, we are interested in the system

[ a11(x0) a12(x0)a21(x0) a22(x0) ][ D1h1(y0)

D1h2(y0) ] = [ 10]

Because we are given thata11(x0)a22(x0)>a12(x0)a21(x0), a21(x0)>0, anda22(x0)>0, we can use Cramer’s Rule to solve this system forD1h1(y0) andD1h2(y0), whichgives the desired inequalities:

D1h1(y0) = ∣ 1 a12(x0)0 a22(x0) ∣

∣ a11(x0) a12(x0)a21(x0) a22(x0) ∣

=a22(x0)

a11(x0)a22(x0)−a12(x0)a21(x0) > 0

D1h2(y0) = ∣ a11(x0) 1a21(x0) 0

∣∣ a11(x0) a12(x0)

a21(x0) a22(x0) ∣=

−a21(x0)a11(x0)a22(x0)−a12(x0)a21(x0) < 0

Problem 29(Implicit Function Theorem).

Let X be an open set inR3, and letf be a continuously differentiable function fromX toR.

Let (x1, x2, x3) be a point inX, such that (i)f (x1, x2, x3) = 0, and (ii) for eachi ∈ {1,2,3},Di f (x1, x2, x3) ≠ 0.

(a) Use the implicit function theorem to obtain three open sets A1,A2

,A3, each inR2

, containing(x2, x3),(x1, x3), and(x1, x2) respectively, and three open setsB1,B2

,B3,each inR, containing ¯x1, x2, and x3 respectively, and unique functionsgi ∶ Ai

→ Bi (fori ∈ {1,2,3}), such that:

(i) f (g1(x2,x3),x2,x3) = 0 for all (x2,x3) ∈A1(ii) f (x1,g2(x1,x3),x3) = 0 for all (x1,x3) ∈A2(iii ) f (x1,x2,g3(x1,x2)) = 0 for all (x1,x2) ∈A3

and:x1 = g1(x2, x3), x2 = g2(x1, x3), x3 = g3(x1, x2)

Further,gi is continuously differentiable onAi for i ∈ {1,2,3}.(b) Using (a), show that:

D1g1(x2, x3)D2g2(x1, x3)D1g3(x1, x2) = −1


Solution.

(a) To solve this problem, we need to apply the Implicit FunctionTheorem three times.It is enough to work through only one of these three cases and note that the othertwo are very similar.

Claim: There is an open setA1 ⊂R2 containing(x2, x3), an open setB1 ⊂R contain-ing x1, and a unique functiong1 ∶A1

→B1 such that:

(1) f (g1(x2,x3),x2,x3) = 0 for all (x2,x3) ∈A1

(2) x1 = g1(x2, x3)Further,g1 is continuously differentiable onA1.

Proof: To establish the claim, we want to apply the Implicit Function Theorem tof at the point(x1,(x2, x3)) ∈ X. Note thatx1 is treated as a variable andx2 andx3

are treated as parameters. We need to check some conditions before applying thetheorem:

● The functionf is defined on the setX ⊂R3. We are told thatX is open and thatf is continuously differentiable onX.

● We are given thatf (x1,(x2, x3)) = 0.

● We are given thatD1 f (x1,(x2, x3)) ≠ 0.

By the Implicit Function Theorem, then, the claim above holds.

(b) From part (a) we have thatf (g1(x2,x3),x2,x3) = 0 for all (x2,x3) ∈ A1. Using theChain Rule to differentiate this with respect tox2, we have

D1 f (g1(x2,x3),x2,x3)D1g1(x2,x3)+D2 f (g1(x2,x3),x2,x3)(1)+D3 f (g1(x2,x3),x2,x3)(0) = 0

Evaluating at(x1, x2, x3) and using the fact that ¯x1 = g1(x2, x3), this becomes

D1 f (x1, x2, x3)D1g1(x2, x3)+D2 f (x1, x2, x3) = 0

Because we are given thatD1 f (x1, x2, x3) ≠ 0, we can rearrange this for

D1g1(x2, x3) = −D2 f (x1, x2, x3)D1 f (x1, x2, x3) (4.1)


From part (a) we have thatf (x1,g2(x1,x3),x3) = 0 for all (x1,x3) ∈ A2. Using theChain Rule to differentiate this with respect tox3, we have

D1 f (x1,g2(x1,x3),x3)(0)+D2 f (x1,g2(x1,x3),x3)D2g2(x1,x3)+D3 f (x1,g2(x1,x3),x3)(1) = 0


D2 f (x1, x2, x3)D2g2(x1, x3)+D3 f (x1, x2, x3) = 0


D2g2(x1, x3) = −D3 f (x1, x2, x3)D2 f (x1, x2, x3) (4.2)

From part (a) we have thatf (x1,x2,g3(x1,x2)) = 0 for all (x1,x2) ∈ A3. Using theChain Rule to differentiate this with respect tox1, we have

D1 f (x1,x2,g3(x1,x2))(1)+D2 f (x1,x2,g3(x1,x2))(0)+D3 f (x1,x2,g3(x1,x2))D1g3(x1,x2) = 0


D1 f (x1, x2, x3)+D3 f (x1, x2, x3)D1g3(x1, x2) = 0


D1g3(x1, x2) = −D1 f (x1, x2, x3)D3 f (x1, x2, x3) (4.3)

Combining equations (4.1), (4.2), and (4.3), we have the desired result:

D1g1(x2, x3)D2g2(x1, x3)D1g3(x1, x2) = −1


ReferencesThis material is based onThe Elements of Real Analysisby Robert Bartle(Chapter

7); Mathematical Analysis: A Modern Approach to Advanced Calculus by Tom Apostol(Chapters 6, 7) andCalculus on Manifoldsby Michael Spivak(Chapter 2).

Chapter 7

Convex Analysis

7.1 Convex Sets

Line SegmentIf x,y ∈Rn, the line segmentjoining x andy is given by the set of points{z ∈Rn ∶ z=

θx+(1−θ)y for some 0≤ θ ≤ 1}.Convex Set

A set S⊂ Rn is a convex setif for every x,y ∈ S, the line segment joiningx andy iscontained inS.

For example, the set of points{(x1,x2) ∈R2 ∶ x21+x2

2 ≤ 1} is a convex set. The set ofpoints{(x1, x2) ∈R2 ∶ x2

1+x22 = 1} is not a convex set.

It can be checked that if two setsS1 andS2 are convex sets inRn, then(i) the intersection ofS1 andS2 [that is, the set{z∈ Rn ∶ z ∈ S1 andz∈ S2}] is a

convex set inRn.(ii) the sum ofS1 andS2 [that is, the set{z∈Rn ∶ z= x+y,wherex∈S1and y∈S2}]

is a convex set inRn.(iii) the (Cartesian) product ofS1 andS2 [that is the set{z∈R2n ∶ z= (x,y), where

x ∈S1 andy ∈S2}] is a convex set inR2n.However, ifS1 andS2 are convex sets inRn, it doesnot follow that the union ofS1 and

S2 [that is, the set{z∈Rn ∶ z∈S1 or z∈S2}] is a convex set inRn. For example, the intervalS1 ≡ [0,1] is a convex set inR, and so is the interval,S2 ≡ [3,4]. The point 2 is on the linesegment joining 1 and 3, but 2 is not in the union of the setsS1 andS2. So the union ofS1

andS2 is not a convex set inR.A vectory ∈Rn is said to be aconvex combinationof the vectorsx1

, ...,xm ∈Rn if thereexistm non-negative real numbersθ1, ...,θm such that

107

CHAPTER 7. CONVEX ANALYSIS 108

(i) m∑i=1

θi = 1; and (ii) y=m∑

i=1θixi

A convex setA⊂Rn can be redefined as a set such that for everytwo vectors in the setA,all convex combinations of thesetwovectors are also in the setA. It can be shown that inthe above statement “two” can be replaced by “m” wherem is any integer exceeding one.

Proposition 3. A set A⊂Rn is convex if and only if for every integer m> 1, and every mvectors in A, every convex combination of the m vectors is in A.

7.2 Separating Hyperplane Theorem for Convex Sets

Hyperplane:Let p∈Rn with p≠0, and letα ∈R. The setH = {x∈Rn ∶ px=α} is called ahyperplane

in Rn with normal p.A hyperplaneH in Rn dividesRn into the two sets:S1 ≡ {x ∈Rn ∶ px≥ α} andS2 ≡ {x ∈Rn ∶ px≤ α}The setsS1 andS2 are called theclosed half-spacesassociated with the hyperplaneH.A very important resu1t on convex sets can now be stated.

Theorem 19. (Minkowski Separation Theorem)Let X and Y be non-empty convex sets inRn, such that X is disjoint from Y. Then there

exists p∈Rn, ∥p∥ = 1, andα ∈R, such that

px ≥ α f orall x ∈X

py ≤ α f orall y ∈Y

The Minkowski separation theorem can be used to establish a criterion for the existenceof non-negativesolutions to a system of linear equations.

Theorem 20. (Farkas Lemma)Exactly one of the following alternatives holds. Either the equation

A(m×n)

x(n×1)

= b(m×1)

(7.1)

has a non-negative solution; or the inequalities

y(1×m)

A(m×n)

≥ 0; y(1×m)

b(m×1)

< 0 (7.2)

has a solution.


Proof. First, suppose that (7.1) doesnot have a non-negative solution. Define

Q= {q ∈Rn ∶ q= n∑i=1

λiAi f or some(λ1, ...,λn) ≥ 0}It can be checked thatQ is a closed convex set. By our hypothesisb is not in Q. SinceQ is closed, there is an open ballB(b,r) ⊂∼Q. That is,Q andB(b,r) are disjoint. SinceB(b,r) is clearly convex, we can use Theorem 19 to obtainp∈Rn, ∥p∥ = 1, andα ∈R suchthat

pq≤ α f orall q ∈B(b,r) (7.3)

andpq≥ α f orall q ∈Q (7.4)

Note that ifq ∈Q, thentq ∈Q for everyt > 0. Using this in (7.4), forq ∈Q,

pq≥ (α/t) f orevery t> 0 (7.5)

Clearly (7.5) implies thatpq≥ 0 f orall q ∈Q (7.6)

SinceAi ∈Q for i = 1, ...,n, so (7.6) implies

pA≥ 0 (7.7)

Using (7.4) again, we note that since 0∈Q, we haveα ≤ 0. Using this in (7.3), wehave

pq≤ 0 f orall q ∈B(b,r) (7.8)

Defineq∗ = b+(r/2)p. Then∥q∗−b∥ = ∥(r/2)p∥ = (r/2)∥p∥ = (r/2). Soq∗ ∈B(b,r), andby (7.8),

pb+(r/2) = pb+(r/2)∥p∥2 = pq∗ ≤ 0

This implies thatpb< 0 (7.9)

Now, (7.7) and (7.9) show that we have demonstrated a solution to the inequalities givenin (7.2).

To complete the proof of Theorem 20, consider, next, that (7.1) has a non-negativesolution, sayx ∈Rn

+. We have to show that (7.2) does not have a solution. Suppose (7.2)did have a solution, sayy ∈Rn, then

0≤ (yA)x= y(Ax) = yb< 0

which is clearly a contradiction.


7.3 Continuous and Differentiable Functions on ConvexSets

We now provide three very useful theorems on continuous and differentiable functions onconvex sets. They are known as the Intermediate Value theorem, the Mean Value theoremand Taylor’s theorem.

Theorem 21. (Intermediate Value Theorem):Suppose A is a convex subset ofRn, and f ∶ A→ R is a continuous function on A.

Suppose x1 and x2 are in A, and f(x1) > f (x2). Then given any c∈R such that f(x1) > c>f (x2), there is0< θ < 1 such that f[θx1+(1−θ)x2] = c.

Example:SupposeX ≡ [a,b] is a closed interval inR (with a < b). Supposef is a continuous

function onX.We know, by Weierstrass theorem, that there will existx1 andx2 in X such thatf (x1) ≥

f (x) ≥ f (x2) for all x ∈X. If f (x1) = f (x2) [this is the trivial case], thenf (x) = f (x1) forall x ∈X, and sof (X) is the single point,f (x1).

If f (x1) > f (x2), then using the fact thatX is aconvex set, we can conclude from theIntermediate Value Theorem that every value betweenf (x1) and f (x2) is attained by thefunction f at some point inX. In other words,f (X) is itself a closed interval.

Theorem 22. (Mean Value Theorem)Suppose A is an open convex subset ofRn, and f ∶A→R is continuously differentiable

on A. Suppose x1 and x2 are in A. Then there is0≤ θ ≤ 1 such that

f (x2)− f (x1) = (x2−x1)∇ f (θx1+(1−θ)x2)Example:

Let f ∶R→R be a continuously differentiable function with the property that f ′(x) > 0for all x∈R. Then given anyx1

, x2 in R, with x2 > x1 we have by the Mean-Value Theorem(sinceR is open and convex), the existence of 0≤ θ ≤ 1, such that

f (x2)− f (x1) = (x2−x1) f ′(θx1+(1−θ)x2)Now f ′(θx1+(1−θ)x2) > 0 by assumption, andx2 > x1 by hypothesis. Sof (x2) > f (x1).This shows thatf is anincreasing functiononR.

A word of caution: a functionf ∶R→R can be increasing without satisfyingf ′(x) > 0at allx ∈R. For example,f (x) = x3 is increasing onR, but f ′(0) = 0.


Theorem 23. (Taylor’s Expansion up to Second-Order)Suppose A is an open, convex subset ofRn, and f ∶A→R is twice continuously differ-

entiable on A. Suppose x1 and x2 are in A. Then there exists0≤ θ ≤ 1, such that

f (x2)− f (x1) = (x2−x1)∇ f (x1)+ 12(x2−x1)H f (θx1+(1−θ)x2)(x2−x1)

7.4 Concave Functions

Let A be a convex set inRn. Then f ∶ A→ R is a concave function(on A) if for all x1,x2 ∈A, and for all 0≤ θ ≤ 1,

f [θx1+(1−θ)x2] ≥ θ f (x1)+(1−θ) f (x2)The functionf is strictly concaveonA if f [θx1+(1−θ)x2] > θ f (x1)+(1−θ) f (x2) when-everx1, x2 ∈A, x1 ≠ x2 and 0< θ < 1.

The relation between concave functions and convex sets is given by the followingresult, which can be proved easily from the definitions of a convex set and a concavefunction.

Theorem 24. Suppose A is a convex subset ofRn and f is a real-valued function on A.Then f is a concave function if and only if the set{(x,α) ∈A×R ∶ f (x) ≥ α} is a convex setin Rn+1.

The following result on concave functions is also useful, although it does not providea characterization of concave functions.

Theorem 25. Suppose A is a convex subset ofRn, and f ∶ A→ R is a concave function.Then, for everyα ∈R, the set

S(α) = {x ∈A ∶ f (x) ≥ a}is a convex set inRn.

A result on concave functions which parallels Proposition 3on convex sets can now benoted. It is known as Jensen’s inequality, and is a very useful tool in convex analysis.

Proposition 4. (Jensen’s Inequality)Let A be a convex subset ofRn

,and f a real-valued function on A. Then a necessaryand sufficient condition for f to be concave is that for each integer m> 1,

f ( m∑i=1

θixi) ≥ m∑i=1

θi f (xi)whenever x1, ...,xm ∈A, (θ1, ...,θm) ∈Rm

+ andm∑

i=1θi = 1.


In general, ifA is a convex set inRn, and f ∶A→R is concave onA, then f need not becontinuous onA. For example, supposeA=R+, and f ∶R+→R is defined by:f (x) = 1+xfor x> 0; f (x) = 0 for x= 0. Then f is a concave function onA, but f is not continuous atx= 0.

If A is anopenconvex set inRn, and f ∶A→R is concave onA, then one can show thatf is continuous onA.

Theorem 26. Suppose A is an open convex subset ofRn, and f ∶ A→ R is a concavefunction on A. Then f is a continuous function on A.

Differentiable Concave Functions:If A ⊂ Rn is an open, convex set, andf ∶ A→ R is continuously differentiable onA,

then we can find a convenient characterization forf to be concave onA in terms of acondition which involves the gradient vector off . (This is particularly useful in concaveprogramming).

Theorem 27. Suppose A⊂Rn is an open set, and f∶A→R is continuously differentiableon A. Then f is concave on A if and only if

f (x2)− f (x1) ≤∇ f (x1)(x2−x1)whenever x1 and x2 are in A.

Corollary 4. Suppose A⊂Rn is an open, convex set, and f∶A→R is continuously differ-entiable on A. Then f is concave on A if and only if

[∇ f (x2)−∇ f (x1)][x2−x1] ≤ 0

whenever x1 and x2 are in A.

It is interesting to note that a characterization ofstrictly concave functions can be givenby replacing the weak inequalities in Theorem 27 and Corollary 4 with strict inequalities(for x1

, x2 in A with x1 ≠ x2).

Theorem 28. Suppose A⊂Rn is an open, convex set, and f∶A→R is continuously differ-entiable on A. Then f is strictly concave on A if and only if

f (x2)− f (x1) <∇ f (x1)(x2−x1)whenever x1, x2 ∈A and x1 ≠ x2.


Corollary 5. Suppose A⊂Rn is an open convex set, and f∶ A→R is continuously differ-entiable on A. Then f is strictly concave on A if and only if

[∇ f (x2)−∇ f (x1)][x2−x1] < 0

whenever x1, x2 ∈A and x1 ≠ x2.

Twice-Differentiable Concave FunctionsIf A⊂Rn is an open set, andf ∶ A→R is twice continuously differentiable onA, then

we can find a convenient characterization forf to be a concave function in terms of thenegative semi-definiteness of the Hessian matrix off .

Theorem 29. Suppose A⊂Rn is an open, convex set, and f∶A→R is twice continuouslydifferentiable on A. Then f is concave on A if and only if Hf (x) is negative semi-definitewhenever x∈A.

If the Hessian off is actually negative definite for allx ∈ A, then f is strictly concaveonA; but the converse is not true.

Theorem 30. Suppose A⊂Rn is an open, convex set, and f∶A→R is twice continuouslydifferentiable on A. If Hf (x) is negative definite for every x∈A, then f is strictly concaveon A.

Example: Let f ∶ R→ R be definied byf (x) = −x4 for all x ∈ R. This is a twice con-tinuously differentiable function on the open, convex setR. It can be checked thatf isstrictly concave onR, but sincef ′′(x) = −12x2, f ′′(0) = 0. This shows that the converseof Theorem 29 is not valid.

7.5 Quasi-Concave Functions

Let A⊂Rn be a convex set, andf a real-valued function onA. Then f is quasi-concaveonA if

f (x2) ≥ f (x1) implies f[θx1+(1−θ)x2] ≥ f (x1)wheneverx1, x2 ∈ A, and 0≤ θ ≤ 1. The function f is strictly quasi-concaveon A iff (x2) ≥ f (x1) implies f [θx1+ (1−θ)x2] > f (x1) wheneverx1, x2 ∈ A, with x1 ≠ x2, and0< θ < 1.

While the condition stated in Theorem 25 did not characerize concave functions, itdoescharacterize quasi-concave functions.


Theorem 31. Suppose A is a convex subset ofRn, and f is a real-valued function on A.Then f is quasi-concave on A if and only if for everyα ∈R, the set

S(α) = {x ∈A ∶ f (x) ≥ α}is a convex set inRn.

Using the concept of strict quasi-concavity, one can provide the following result on theuniqueness of solutions to constrained maximization problems.

Theorem 32. (Uniqueness of Solutions)Suppose A is a non-empty, compact and convex set inRn. Suppose f∶ A→ R is a

continuous, strictly quasi-concave function on A. Then, there exists x1 ∈ A such that forall x ∈A which are not equal to x1, we have f(x) < f (x1).Proof. By Weierstrass theorem, there isx1 ∈A such that

f (x) ≤ f (x1) f orall x ∈A (7.10)

If the claim of the Theorem were not true, there would exist somex2 ∈A, x2 ≠ x1, such thatf (x2) = f (x1). But thenx3 ≡ [(1/2)x1+(1/2)x2] would belong toA (sinceA is a convexset), and by strict quasi-concavity off onA, we would havef (x3)= f [(1/2)x1+(1/2)x2]>f (x1), which contradicts (7.10).

Differentiable Quasi-Concave Functions:A characterization of differentiable quasi-concave functions can be given which paral-

lels the characterization of differentiable concave functions stated in Theorem 27. (Thisis particularly useful in Quasi-Concave Programming).

Theorem 33. Suppose A⊂Rn is an open, convex set, and f∶A→R a continuously differ-entiable function. Then f is a quasi-concave if and only if

f (x2) ≥ f (x1) implies(x2−x1)∇ f (x1) ≥ 0

whenever x1, x2 ∈A.

Twice Differentiable Quasi-Concave FunctionsAn interesting characterization of twice continuously differentiable quasi-concave func-

tions can be given in terms of the “bordered” Hessian matrix associated with the functions.Let A be an open subset ofRn, and f ∶ A→ R be a twice continuously differentiable

function onA. Thebordered Hessian matrix of fat x ∈ A is denoted byGf (x) and isdefined as the following(n+1)×(n+1)matrix

Gf (x) = [ 0 ∇ f (x)∇ f (x) H f (x) ]We denote the(k+1)th leading principal minor ofGf (x) by ∣Gf (x;k)∣, wherek= 1, ...,n.


Theorem 34. Let A be an open convex set inRn, and f ∶ A→R be a twice continuouslydifferentiable function on A.

(i) If f is quasi-concave on A, then(−1)k ∣Gf (x;k)∣ ≥ 0 for x ∈A, and k= 1, ...,n.(ii) If (−1)k ∣Gf (x;k)∣ > 0 for x ∈A, and k= 1, ...,n, then f is strictly quasi-concave

on A.



Problem 30(Intermediate Value Theorem).

Here is a statement of the Intermediate Value Theorem for continuous real valuedfunctions of a real variable.

Theorem:Let f be a continuous real valued function on the closed intervalA = [a,b]. Suppose

x,y∈A satisfy f (x) > f (y). Then for everyc, satisfying f (x) > c> f (y), there isz∈A, suchthat f (z) = c.

We want to use this theorem to prove the following version of the Intermediate ValueTheorem for continuous real valued functions of several real variables.

Corollary:Let B be a convex subset ofRn

, and letF be a continuous real valued function onB.Supposex1

,x2 ∈ B satisfyF(x1) > F(x2). Then for everyc, satisfyingF(x1) > c> F(x2),there isv ∈B, such thatF(v) = c.

Proceed with the following steps.(a) DefineA= [0,1], and for eacht ∈A, define f (t) = F(tx1+(1− t)x2). This function

is well defined sinceB is a convex set, andF is defined onB. Verify that f is a continuousfunction onA.

(b) Note thatf (1) = F(x1) > F(x2) = f (0). Use the Theorem (stated above) to obtainz∈A such thatf (z) = c. This meansF(zx1+(1−z)x2) = c.

(c) Definev= zx1+(1−z)x2, and verify that this proves the Corollary.

Solution.

(a) Note that an arbitraryx1 ∈ B andx2 ∈ B are given in the hypothesis of the Corollary.Now, letA= [0,1]. Define f ∶A→R by

f (t) = F(tx1+(1− t)x2) for all t ∈A

Note that f is well-defined: sincex1 ∈ B andx2 ∈ B, by convexity ofB we have thattx1+(1− t)x2 ∈B for all t ∈A.

To show thatf is continuous onA, fix some t ∈ A and someε > 0. We want toshow that there is someδ > 0 such that whenevert ∈ A and d(t, t) < δ, we have∣ f (t)− f (t)∣ < ε. Note that sinceA⊂R, we can writed(t, t) = ∣t − t ∣. Also, note that

∣ f (t)− f (t)∣ = ∣F(tx1+(1− t)x2)−F(tx1+(1− t)x2)∣


To show thatf is continuous at our arbitraryt ∈ A, we can use the fact thatF iscontinuous at the pointtx1+ (1− t)x2 ∈ B. That is, givenε > 0 (note that this is thesameε as above), by continuity ofF there is someδ′ > 0 such that wheneverx ∈ Bandd(x, tx1+(1− t)x2) < δ′, we have∣F(x)−F(tx1+(1− t)x2)∣ < ε.

This is looking similar to the inequality we want to establish. Now, recall that foranyy,z,w ∈Rn and anyλ ∈R, we have thatd(y+w,z+w) = d(y,z) andd(λy,λz) =∣λ∣ d(y,z). With the goal of picking an appropriateδ, consider the following dis-tance:

d(tx1+(1− t)x2, tx1+(1− t)x2) = d(tx1− tx1

,(1− t)x2−(1− t)x2)= d((t − t)x1

,(t − t)x2)= ∣t − t ∣ d(x1

,x2)Now, letδ = δ′

d(x1,x2) > 0. Whenevert ∈A and∣t− t ∣ < δ, we have thattx1+(1−t)x2 ∈B

and

d(tx1+(1− t)x2, tx1+(1− t)x2) = ∣t − t ∣ d(x1

,x2)< δ d(x1

,x2)= δ′

Then by continuity ofF we have that∣F(tx1+(1− t)x2)−F(tx1+(1− t)x2)∣ < ε,which is equivalent to∣ f (t)− f (t)∣ < ε. Sincet was chosen arbitrarily, this showsthat f is continuous onA.

(b) By the definition off , f (1) = F(x1) and f (0) = F(x2). By hypothesis in the Corol-lary, F(x1) > F(x2), which is equivalent tof (1) > f (0). Let some scalarc satisfyf (1) > c> f (0). Then sincef is continuous on the closed intervalA, we have by thestated Theorem that there is somez∈A such thatc= f (z) = F(zx1+(1−z)x2).

(c) Definev = zx1+ (1−z)x2. SinceB is convex andz∈ A = [0,1], we have thatv ∈ B.Using part (b), we have that for anyc satisfyingF(x1) > c > F(x2), there is somev ∈B such thatF(v) = c. This proves the Corollary.

Problem 31(Mean Value Theorem).

Here is a statement of the Mean Value Theorem for real valued functions of a real variable.Theorem:


Let f be a continuous real valued function on the closed intervalA= [a,b]. Supposefis differentiable for allx ∈ (a,b). Then there isc ∈ (a,b) such that:

f (b)− f (a) = (b−a) f ′(c)Use this theorem to prove the following version of the Mean Value Theorem for real

valued functions of several real variables.Corollary:Let B be an open convex subset ofRn

, and letF be a continuously differentiable realvalued function onB. Supposex1

,x2 ∈B. Then there isθ ∈ (0,1), satisfying:

F(x2)−F(x1) = (x2−x1)∇F(θx1+(1−θ)x2)Solution.

We can use the same functionf that we used in Problem 1 to help us establish thisresult. Note that an arbitraryx1 ∈B andx2 ∈B are given in the hypothesis of the Corollary.Now, letA= [0,1]. Define f ∶A→R by

f (t) = F(tx1+(1− t)x2) for all t ∈A

We showed in part (a) of problem 1 thatf is well-defined and continuous onA. Now,because we have made the additional assumptions in this problem thatB is open andFis continuously differentiable onB, we can use the Chain Rule to differentiatef on theopen interval(0,1). Remember thatx1

i denotes theith component of the vectorx1, andx2i

denotes theith component of the vectorx2. For all t ∈ (0,1), we have

f ′(t) = n∑i=1[DiF(tx1+(1− t)x2) ∂

∂t(tx1

i +(1− t)x2i )]

=n∑

i=1[(x1

i −x2i ) DiF(tx1+(1− t)x2)]

= (x1−x2) ∇F(tx1+(1− t)x2)SinceF is continuously differentiable onB andtx1+(1− t)x2 ∈ B whenevert ∈ (0,1), wehave thatf ′(t) is continuous on(0,1). By the stated Theorem, there isθ ∈ (0,1) such thatf (1)− f (0) = (1−0) f ′(θ). And by the definition off , we have thatf (1) = F(x1) andf (0) = F(x2). Using the above expression forf ′, then, we have that for someθ ∈ (0,1),

F(x1)−F(x2) = (x1−x2) ∇F(θx1+(1−θ)x2)Multiplying each side of this equation by−1 gives the desired result.


Problem 32(Jensen’s Inequality for Concave Functions).

SupposeA is a convex subset ofRn, and f a concave function onA. Let x1

,x2,x3 ∈ A,

and letθ1,θ2,θ3 be real numbers satisfyingθi ≥ 0 for i = 1,2,3, and(θ1+θ2+θ3) = 1.(a) Using the definition of a convex set, verify that:

(θ1x1+θ2x2+θ3x3) ∈A

(b) Using the definition of a concave function, show that:

f (θ1x1+θ2x2+θ3x3) ≥ θ1 f (x1)+θ2 f (x2)+θ3 f (x3)Solution.

(a) If θ1 = 0, θ2 = 0, orθ3 = 0 the problem is trivial, becauseθ1x1+θ2x2+θ3x3 becomesa convex combination of just two vectors inA, and we know thatA is convex. So,assume thatθ1 ≠ 0, θ2 ≠ 0, andθ3 ≠ 0. Then we can write

θ1x1+θ2x2+θ3x3 = (θ1+θ2) θ1

θ1+θ2x1+(θ1+θ2) θ2

θ1+θ2x2+θ3x3

We know that θ1θ1+θ2

≥ 0, θ2θ1+θ2

≥ 0, and θ1θ1+θ2

+ θ2θ1+θ2

= 1. Then sinceA is convex,

x=θ1

θ1+θ2x1+ θ2

θ1+θ2x2 ∈A

Again, sinceθ1+θ2 ≥ 0, θ3 ≥ 0, andθ1+θ2+θ3 = 1, convexity of A implies that

(θ1+θ2)x+θ3x3 ∈A

This is equivalent toθ1x1+θ2x2+θ3x3 ∈A, which is what we wanted to show.

(b) As in part (a), ifθ1 = 0, θ2 = 0, or θ3 = 0 the result will follow directly from theconcavity of f onA. So, assume thatθ1 ≠ 0, θ2 ≠ 0, andθ3 ≠ 0. We can get the resultby twice applying the concavity off on A. Note that the exact conditions we needto verify before applying concavity have been shown in part (a) to be satisfied.

f ((θ1+θ2)x+θ3x3) ≥ (θ1+θ2) f (x)+θ3 f (x3)= (θ1+θ2) f ( θ1

θ1+θ2x1+ θ2

θ1+θ2x2)+θ3 f (x3)

≥ (θ1+θ2)[ θ1

θ1+θ2f (x1)+ θ2

θ1+θ2f (x2)]+θ3 f (x3)

= θ1 f (x1)+θ2 f (x2)+θ3 f (x3)


Problem 33(Test for Concavity).

Let f ∶R2+→R be defined by:

f (x,y) =Axayb for all (x,y) ∈R2+

whereA, a andb are positive parameters. [It is known thatf is continuous onR2+ andtwice continuously differentiable onR2++. ]

(a) Show thatf is concave onR2+ if (a+b) ≤ 1.(b) Show thatf is not concave onR2+ if (a+b) > 1.

Solution.

(a) Testing for concavity by testing for negative semi-definiteness of the Hessian offonly works on open sets, but we want to show thatf is concave onR2+, which is notopen. Showing thatf is concave on some open set that includesR2+ will not work,as we will see from the calculations below. So, consider two cases.

Case 1: Sincef is twice continuously differentiable on the open and convexsetR2++,we can show thatf is concave onR2++ by showing that the Hessian off is negativesemi-definite onR2++. The Hessian off at any(x,y) ∈R2++ is

H f (x,y) = [ a(a−1)Axa−2yb abAxa−1yb−1

abAxa−1yb−1 b(b−1)Axayb−2 ]Now, a(a−1)Axa−2yb < 0 andb(b−1)Axayb−2 < 0 for all (x,y) ∈ R2++ becausea ∈(0,1), b ∈ (0,1), andA> 0. The determinant of the Hessian off at any(x,y) ∈R2++is

detH f (x,y) = ab(a−1)(b−1)A2x2a−2y2b−2−a2b2A2x2a−2y2b−2

=A2x2a−2y2b−2ab((a−1)(b−1)−ab)=A2x2a−2y2b−2ab(1−a−b)≥ 0

sincea ∈ (0,1), b ∈ (0,1), anda+b≤ 1. This shows that the Hessian off is negativesemi-definite onR2++, which shows thatf is concave onR2++.

Case 2: Consider(x1,y1) ∈R2+, (x2,y2) ∈R2+, andθ ∈ [0,1]. BecauseR2+ is convex,we know thatθ(x1,y1)+(1−θ)(x2,y2) ∈R2+. Assume that eitherx1 = 0 or y1 = 0, so


that(x1,y1) is on the boundary of the setR2+. Given the specificf we are consider-ing, this means thatf (x1,y1) = 0. Now,

f (θ(x1,y1)+(1−θ)(x2,y2))= f (θx1+(1−θ)x2,θy1+(1−θ)y2)=A(θx1+(1−θ)x2)a(θy1+(1−θ)y2)b≥A((1−θ)x2)a((1−θ)y2)b (sincea> 0 andb> 0)=A(1−θ)a+bxa

2yb2

= (1−θ)a+b f (x2,y2)≥ (1−θ) f (x2,y2) (since 1−θ ∈ [0,1] anda+b≤ 1)= θ f (x1,y1)+(1−θ) f (x2,y2) (since f (x1,y1) = 0)

Combining the two cases, we have thatf is concave onR2+.

(b) Recall from part (a) that the determinant of the Hessian off at any(x,y) ∈R2++ is

detH f (x,y) =A2x2a−2y2b−2ab(1−a−b) < 0

sincea> 0, b> 0, anda+b> 1. This shows that the Hessian off cannot be negativesemi-definite at any point(x,y) ∈R2++. Thereforef is not concave onR2++, so f isnot concave onR2+.

Problem 34(Characterization of Quasi-Concave Functions).

Let f ∶Rn+→R+ be a real valued function.

(a) Show that iff is quasi-concave onRn+, then for everyα ∈R, the set:

U(α) = {x ∈Rn+ ∶ f (x) ≥ α}

is a convex set.(b) Show that if for everyα ∈R, the set:

U(α) = {x ∈Rn+ ∶ f (x) ≥ α}

is a convex set, thenf is quasi-concave onRn+.

Solution.


(a) For all α such thatU(α) is the empty set or has only one element, the result holdstrivially. So, pick someα ∈R such thatU(α) has at least two distinct elements. Letx1,x2 ∈U(α) andθ ∈ [0,1]. We have by the definition ofU(α) that f (x1) ≥ α and

f (x2) ≥ α. Suppose without loss of generality thatf (x1) ≥ f (x2). Since f is quasi-concave onRn

+, we have thatf (θx1+(1−θ)x2)≥ f (x2)≥α, soθx1+(1−θ)x2 ∈U(α).ThusU(α) is a convex set.

(b) Let x1,x2 ∈Rn

+, θ ∈ [0,1], and assume without loss of generality thatf (x1) ≥ f (x2).Since f (x2) ∈R andU(α) is convex for allα ∈R, the set

U( f (x2)) = {x ∈Rn+ ∣ f (x) ≥ f (x2)}

is convex. Now, sincex1 ∈U( f (x2)), x2 ∈U( f (x2)), andU( f (x2)) is convex, wehave thatθx1+(1−θ)x2 ∈U( f (x2)). By the definition ofU( f (x2)), this is equivalentto f (θx1+(1−θ)x2) ≥ f (x2), which shows thatf is quasi-concave onRn

+.

Problem 35(Testing for Quasi-Concavity).

(a) Let f ∶R2+→R be defined by:

f (x,y) =Axayb for all (x,y) ∈R2+

whereA, a andb are positive parameters. [It is known thatf is continuous onR2+ andtwice continuously differentiable onR2++.] Show thatf is quasi-concave onR2+.

(b) Letg ∶R2+→R be defined by:

g(x,y) = xa+yb for all (x,y) ∈R2+

wherea andb are parameters, witha> 1 andb> 1. Show thatg is increasing onR2+ but itis not quasi-concave onR2+.

Solution.

(a) As in Problem 4, part (a), we must consider two cases, sinceR2+ is not open and wewill not be able to show thatf is quasi-concave on some open set that containsR2+.

Case 1: Sincef is twice continuously differentiable on the open and convexsetR2++,we can show thatf is quasi-concave onR2++ by showing that for all(x,y) ∈ R2++,the second leading principal minor of the bordered Hessian of f at (x,y) is strictly


negative and the determinant of the bordered Hessian off at(x,y) is strictly positive.The bordered Hessian off at any(x,y) ∈R2++ is

Bf (x,y) =⎡⎢⎢⎢⎢⎢⎣

0 aAxa−1yb bAxayb−1

aAxa−1yb a(a−1)Axa−2yb abAxa−1yb−1

bAxayb−1 abAxa−1yb−1 b(b−1)Axayb−2

⎤⎥⎥⎥⎥⎥⎦Now, for all (x,y) ∈R2++ the second leading principal minor of this matrix is

∣ 0 aAxa−1yb

aAxa−1yb a(a−1)Axa−2yb ∣ = −a2A2x2a−2y2b < 0

sincea> 0 andA> 0. The determinant of the bordered Hessian off at any(x,y) ∈R2++ is

detBf (x,y) = −aAxa−1yb[ab(b−1)A2x2a−1y2b−2−ab2A2x2a−1y2b−2]+bAxayb−1[a2bA2x2a−2y2b−1−ab(a−1)A2x2a−2y2b−1]= −a2bA3x3a−2y3b−2(b−1−b)+ab2A3x3a−2y3b−2(a−a+1)= a2bA3x3a−2y3b−2+ab2A3x3a−2y3b−2

= abA3x3a−2y3b−2(a+b)> 0

sincea> 0, b> 0, andA> 0. This shows thatf is quasi-concave onR2++.

Case 2: Consider(x1,y1) ∈R2+, (x2,y2) ∈R2+, andθ ∈ [0,1]. BecauseR2+ is convex,we know thatθ(x1,y1)+(1−θ)(x2,y2) ∈R2+. Assume that eitherx1 = 0 or y1 = 0, sothat(x1,y1) is on the boundary of the setR2+. Given the specificf we are consider-ing, this means thatf (x2,y2) ≥ f (x1,y1) = 0. Now,

f (θ(x1,y1)+(1−θ)(x2,y2)) ≥ 0= f (x1,y1)Combining the two cases, we have thatf is quasi-concave onR2+.

(b) Note that sinceg is defined onR2+, which is not open, we can only consider deriva-tives on the open setR2++. For all (x,y) ∈ R2++, we have thatD1g(x,y) = axa−1 > 0sincea> 1 andD2g(x,y) = byb−1 > 0 sinceb> 1. This shows thatg is increasing onR2++. We can check directly thatg is increasing on the boundary. For allx > 0 andy ≥ 0, we have thatg(x, y)−g(0, y) = xa > 0. For all x ≥ 0 andy > 0, we have thatg(x,y)−g(x,0) = yb > 0. Combined with the result thatg is increasing onR2++, wehave thatg is increasing onR2+.


Sinceg is twice continuously differentiable on the open and convexsetR2++, quasi-concavity ofg onR2++ would imply that the determinant of the bordered Hessian ofg at any(x,y) ∈R2++ is non-negative. But we have that at any(x,y) ∈R2++,

detBg(x,y) =RRRRRRRRRRRRRR

0 axa−1 byb−1

axa−1 a(a−1)xa−2 0byb−1 0 b(b−1)yb−2

RRRRRRRRRRRRRR= −axa−1[ab(b−1)xa−1yb−2]+byb−1[−ab(a−1)xa−2yb−1]= −a2b(b−1)x2a−2yb−2−ab2(a−1)xa−2y2b−2

< 0

sincea > 1 andb > 1. Thereforeg is not quasi-concave onR2++, sog is not quasi-concave onR2+.


ReferencesThis material is based onConvex Structures and Economic Theoryby H. Nikaido

(Chapter 1);Mathematical EconomicsbyA. Takayama(Chapters 0,1);Mathematical Anal-ysisby T. Apostol (Chapters 4,6); The Elements of Real Analysisby Robert Bartle(Chap-ters 4,7); andNon-Linear Programmingby O.L. Mangasarian(Chapters 2, 3, 4, 6 and9).

Part III

Classical Optimization Theory

126

Chapter 8

Unconstrained Optimization

8.1 Preliminaries

We will present below the elements of “classical optimization theory”. We will concen-trate on characterizing points ofmaximumof a function of several variables; the theorywhich characterizes points ofminimum of such a function can be inferred without toomuch difficulty.

Our first task will be to look at the theory ofunconstrainedmaximization, and discussthe relevant necessary and sufficient conditions for such anunconstrained maximum tooccur. This is the subject matter of this chapter. Our secondtask will be to present thetheory ofconstrainedmaximization, where the only constraints areequalityconstraints.This is the theory involving the well-known “Lagrange multiplier method”, and is takenup in Chapter 9.Unconstrained Maximization Theory

Our framework is the following. There is a setA⊂Rn; there is a functionf ∶ A→R.We are interested in identifying points inA at which the function attains a (local or global)maximum.Local and Global Maximum

Let A⊂Rn, and let f be a function fromA to R. A point c ∈A is said to be apoint oflocal maximumof f if there existsδ > 0, such thatf (c) ≥ f (x) for all x ∈ A which satisfy∥x−c∥ < δ. It is said to be apoint of global maximumof f if f (c) ≥ f (x) for all x ∈A.

127

CHAPTER 8. UNCONSTRAINED OPTIMIZATION 128

8.2 Necessary Conditions for a Local Maximum

We will present two necessary conditions for a local maximum. One is a condition on thefirst-order partial derivatives of the relevant function (called “first-order conditions”); theother is a condition on the second-order partial derivatives of the relevant function (called“second-order necessary conditions”).

Theorem 35.Let A be an open set inRn, and let f∶A→R be a continuously differentiablefunction on A. If c∈A is a point of local maximum of f , then

∇ f (c) = 0 (8.1)

Remark: The n equations given by (8.1) are called the first-order conditions for a localmaximum.

Theorem 36. Let A be an open set inRn, and let f ∶ A→ R be a twice continuouslydifferentiable function on A. If c∈A is a point of local maximum of f , then

H f (c) is negative semi-definite (8.2)

Remark: The condition (8.2) is called the second-order necessary condition for a localmaximum.

Necessary conditions like (8.1) and (8.2) stated above are useful because they help usto rule out points where a local maximum cannot occur, thereby narrowing our search forpoints where a local maximum does occur. The following two examples illustrate thispoint.Examples:

(i) Let f ∶ R→ R be given byf (x) = 1−x2 for all x ∈ R. ThenR is an open set,and f a continuously differentiable function onR. Consider the pointc= 1. We calculatef ′(c) = f ′(1) = −2(1) = −2. By Theorem 35, we can therefore conclude thatc= 1 is not apoint of local maximum off .

(ii) Let f ∶ R → R be given by f (x) = 1−2x+ x2 for all x ∈ R. ThenR is anopen set, andf a twice continuously differentiable function onR. Consider the pointc= 1. We can calculatef ′(c) = f ′(1) = −2+2(1) = 0, so condition (8.1) of Theorem 35 issatisfied.Notice that Theorem 35, by itself, fails to be of any help at this point. We cannotconclude from Theorem 35 thatc = 1 is a point of local maximum; we cannot concludefrom Theorem 35 thatc = 1 is not a point of local maximum. Theorem 36, however, isuseful at this point. We can calculatef ′′(c) = f ′′(1) = 2 > 0, and so condition (8.2) ofTheorem 36 is violated. Consequently, by Theorem 36, we can conclude thatc= 1 is nota point of local maximum off .


8.3 Sufficient Conditions for a Local Maximum

We present below a set of sufficient conditions for a local maximum.

Theorem 37. Let A be an open set inRn, and let f ∶ A→ R be a twice continuouslydifferentiable function on A. If c∈A, satisfies

∇ f (c) = 0 (8.3)

andH f (c) is negative definite (8.4)

then c is a point of local maximum of f .

Remark: Condition (8.3) of Theorem 37 is called the second-order sufficient conditionfor a local maximum.

It should be noted that condition (8.4) cannot be weakened tocondition (8.2) in thestatement of Theorem 37. The following example illustratesthis point.Example:

Let f ∶ R → R be given by f (x) = x3 for all x ∈ R. ThenR is an open set, andfis a twice continuously differentiable function onR. At c = 0, f ′(c) = f ′(0) = 0, andf ′′(c) = f ′′(0) = 0, so condition (8.3) and condition (8.2) are satisfied. Butc is clearly nota point of local maximum off since f is an increasing function onR.

It may also be observed that condition (8.2) cannot be strengthened to condition (8.4)in the statement of Theorem 36. The following example illustrates this point.Example:

Let f ∶R→R be given byf (x) = −x4 for all x ∈R. ThenR is an open set, andf is atwice continuously differentiable function onR. Clearly,c=0 is a point of local maximumof f [since f (0) = 0, while f (x) < 0 for all x≠ 0]. One can calculate thatf ′(c) = f ′(0) = 0,and f ′′(c) = f ′′(0) = 0. Thus conditions (8.1) and (8.2) are satisfied, but (8.4) isviolated.

The outcome of this discussion is the following: the second-order necessary condi-tions for a local maximum are different from (weaker than) the second-order sufficientconditions for a local maximum. This simply reflects the factthat, in general, the firstand second derivatives of a function at a point do not captureall aspects relevant to theoccurrence of a local maximum of the function at that point.

The sufficient conditions of Theorem 37 enable us to find points of local maximum ofa function, as the following example shows.Example:

Let f ∶R2→R be given byf (x1, x2) = 2x1x2−2x2

1−x22 for all (x1, x2) ∈R2. ThenR2

is an open set, andf is a twice continuously differentiable function. If we write down the


condition (8.1) [or (8.3)] at a point(c1, c2) ∈R2, we get

D1 f (c1, c2) = 2c2−4c1 = 0

D2 f (c1, c2) = 2c1−2c2 = 0

Thus condition (8.3) is satisfied if and only ifc1 = c2 = 0.Next, we can find the Hessian off at (c1, c2) ∈R2:

H f (c1, c2) = [ −4 22 −2 ]

Now,−4< 0, and(−4)(−2)−(2)(2) = 8−4> 0. SoH f (c1, c2) is negative definite for each(c1, c2) ∈R2. Thus, condition (8.4) is clearly satisfied. It follows fromTheorem 37 that(c1, c2) = (0,0) is a point of local maximum off .

8.4 Sufficient Conditions for a Global Maximum

While several sets of sufficient conditioins for a global maximum can be developed, thefollowing two are among the most useful.

Theorem 38. Let A be an open convex set inRn, and let f ∶ A→ R be a continuouslydifferentiable function on A. If c∈A satisfies

∇ f (c) = 0 (8.5)

and f is a concave function on A, then c is a point of global maximum of f .

To see this note that for allx ∈A,

f (x)− f (c) ≤ (x−c)∇ f (c) (8.6)

since f is concave and continuously differentiable onA [See Theorem 9 of Chapter 7 on“Convex Analysis”]. Using (8.5) in (8.6), we getf (x) ≤ f (c) for all x ∈ A, soc is a pointof global maximum off .

Theorem 39. Let A be an open convex set inRn, and let f ∶A→R be a twice continuouslydifferentiable function on A. If c∈A satisfies

∇ f (c) = 0 (8.7)

andH f (x) is negative semi-definite for all x∈A (8.8)

then c is a point of global maximum of f .


To see this, note that (8.8) ensures thatf is concave onA; so the result follows readilyfrom Theorem 38.

It is worth noting that Theorem 38 (or 39) might be applicablein cases where Theorem37 is not applicable. Letf ∶R→R be given byf (x) = −x4. Here, one notes thatf ′(0) = 0and f ′′(x) = −12x2 ≤ 0 for all x. Thus applying Theorem 38 (or 39), we can conclude thatx= 0 is a point of global maximum, and hence of local maximum. But the conclusion thatx= 0 is a point of local maximum cannot be derived from Theorem 37, since f ′′(0) = 0.

8.5 The Method of Least Squares

Suppose we are givenn points(xi , yi), i = 1, ...,n in R2. Let f ∶R→R be given byf (x) =ax+b for all x ∈ R. We wish to find a functionf (that is, we want to choosea ∈ R andb ∈R) such that the quantity

n∑i=1[ f (xi)−yi]2

is minimized.We can set up the problem as anunconstrained maximization problemas follows.

DefineF ∶R2→R by

F(a,b) = − n∑i=1[axi +b−yi]2

The maximization problem then is

Max(a,b)

F(a,b)F is twice continuously differentible onR2, and we can calculate

D1F = −2 n∑i=1[axi +b−yi]xi = −2 n∑

i=1[ax2

i +bxi −xiyi]D2F = −2 n∑

i=1[axi +b−yi]

D11F = −2 n∑i=1

x2i ; D12F = −2 n∑

i=1xi

D21F = −2 n∑i=1

xi; D22F = −2n

Thus, the determinant of the Hessian ofF is

det(HF(a,b)) = 4nn∑

i=1x2

i −4[ n∑i=1

xi]2


Now, by the Cauchy-Schwarz inequality,

∣ n∑i=1

xi∣ ≤ [∑x2i ]1/2n1/2

so

[ n∑i=1

xi]2 ≤ n∑x2i

and consequently, det(HF(a,b))≥0. SinceD11F(a,b)≤0,D22F(a,b)≤0, and det(HF(a,b))≥0,HF(a,b) is negative semi-definite. Consequently, if(a∗, b∗) satisfies the first-orderconditions, then(a∗, b∗) is a point of global maximum off by Theorem 39. The first-order conditions are

an∑

i=1x2

i +bn∑

i=1xi =

n∑i=1

xiyi (8.9)

an∑

i=1xi +bn=

n∑i=1

yi (8.10)

Denoting(1/n) n∑i=1

xi by x and(1/n) n∑i=1

yi by y [the means ofx andy respectively], we get

from (8.10) thatax+b= y (8.11)

Using this in (8.9) leads to

an∑

i=1x2

i +(y−ax)nx=n∑

i=1xiyi (8.12)

Thus,

(1/n) n∑i=1

xiyi −xy

(1/n) n∑i=1

x2i −x2

= a

y−ax = b

solves the problem, provided not all thexi are the same.


8.6 The Envelope Theorem

Let X be an open convex set ofRn, andA be an open subset ofRm. Let f be a twicecontinuously differentiable function onX ×A which is quasi-concave onX, givena ∈ A.We interpret(x1, ...,xn) as the “variables” and(a1, ...,am) as the “parameters”.

Givena ∈A, we can formulate the following maximization problem:

Maxx∈X

f (x;a) (P)

Let a∗ ∈A. Supposex ∈X is a point such that

∇ f (x;a∗) = 0

H f (x;a∗)is negative definite

Using Theorem 37, we then know thatx is a point of local maximum off , givena∗. Givena∗ ∈A, f is quasi-concave onX, sox can be shown to be a solution of problem(P), givena∗. (Can you show this?) In fact, ˆx can be shown to be the unique solution to(P), givena∗.

Notice that by the implict function theorem, there is an opensetB containinga∗, andan open setC containingx, and a unique functiong ∶B→C, such that

(i) x= g(a∗)(ii) ∇ f [g(a);a] = 0 f ora ∈B

Also, g is continuously differentiable onB. Furthermore, one can choose the open setBso as to ensure

(iii) H f [g(a;a] is negative definite fora ∈BFor a ∈ B, [a not necessarily equal toa∗], we then note by (ii) and (iii), thatg(a) is a

point of local maximum off , given a. And sincef is quasi-concave onX [given a ∈ B],g(a) is the unique solution to problem (P), givena.

We, can now define for problem (P), thevalue, V ∶B→R by

V(a) =Maxx∈X

f (x;a)Furthermore, we can define themaximizer h∶B→X by

h(a) = {z∈X ∶ f (z;a) ≥ f (x;a) f orall x ∈X}What we have just established in the previous paragraph is that

V(a) = f [g(a);a] f ora ∈B (8.13)


h(a) = g(a) f ora ∈B (8.14)

Two questions which now arise quite naturally are:(1) If a changes a little froma∗, how will the maximized value off change?(2) If a changes a little froma∗, how will the maximizer change?

Suppose, for concreteness, we want to answer these questions for a parameter change,where onlya1 changes. Then question (1) can be answered by finding outD1V(a∗); andquestion (2) can be answered by finding out the vector[D1g1(a∗),D1g2(a∗), ....,D1gn(a∗)].

Using (8.13), and the chain-rule,

D1V(a∗) = n∑i=1

Di f [g(a∗); a∗]D1gi(a∗)+Dn+1 f [g(a∗); a∗]Using (ii), we haveDi f [g(a∗); a∗] = 0for i = 1, ...,n. Hence,

D1V(a∗) =Dn+1 f [x; a∗] (8.15)

This result is known as the “envelope theorem”, and it answers question (1) above.To answer question (2), use (ii) to obtain (employing again the chain-rule),

n∑i=1

D1i f [g(a∗);a∗]D1gi(a∗)+D1n+1 f [g(a∗);a∗] = 0

−−−−−−−−−−−−−−−−−−−−−−−−−n∑

i=1Dni f [g(a∗);a∗]D1gi(a∗)+Dnn+1 f [g(a∗);a∗] = 0

Then using (iii), we can employ Cramer’s Rule to obtain the vector [D1g1(a∗), ...,D1gn(a∗)].This answers question (2).

As an application, considerX ≡R++, andA=R2++, with f ∶X×A→R given by

f (x; p,w) = pφ(x)−wx

We interpretφ ∶R+→R as the production function [withx as the input level andφ(x) thecorresponding output level],p as the output price andw as the input price. Thus,f (x; p,w)is the profit ifx of input is employed when the output price isp, and input price isw.

Assumingφ(x) = 2x1/2 for x≥ 0, we observe thatf is twice continuously differentiableonX×A, and given(p,w) ∈A, f is quasi-concave onX, sinceφ is concave onX.

Let (p∗,w∗) ∈A. Then, we note that

0=D1 f (x; p∗,w∗) = (p∗/x1/2)−w∗


whenx= (p∗/w∗)2. Furthermore,

D11 f (x; p∗,w∗) = −[p∗/2x3/2] < 0

Then applying the implicit function theorem, we can obtain an open setB containing(p∗,w∗) and an open setC containingx, and a unique functiong ∶B→C, such that(i) x= g(p∗,w∗)(ii) D1 f [g(p,w); p,w] = 0 f or(p,w) ∈B

Further,g is continuously differentiable onB.Thevaluefor this problem is known as the “profit function”, and by (8.13), it is given

byπ(p,w) = f [g(p,w); p,w]

The maximizer is known as the (input)demand function, and by (8.14), it is given by

x(p,w) = g(p,w)If we answer question (1) in this framework, we get, from (8.15),

D1π(p∗,w∗) = φ(g(p∗,w∗)) = φ(x(p∗,w∗))D2π(p∗,w∗) = −g(p∗,w∗) = −x(p∗,w∗)

This is generally known as “Hotelling result” in the theory of the firm.



Problem 36(Unconstrained Optimization: First-Order Conditions).

Here is a statement of the first-order condition for a maximumof real valued functionsof a real variable.

Theorem:Let f be a continuously differentiable real valued function on the intervalA= (a,b). If

c ∈A satisfiesf (c) ≥ f (t) for all t ∈A, then f ′(c) = 0.We want to use this theorem to prove the following version of the first-order condition

for a local maximum of real valued functions of several real variables.Corollary:Let C be an open subset ofRn, and letF be a continuously differentiable real valued

function onC. If x ∈C is a point of local maximum ofF , then∇F(x) = 0.Proceed with the following steps.(a) Since ¯x ∈C is a point of local maximum ofF , we can findr > 0 such thatB ≡

B(x,2r) ⊂C, andF(x) ≥ F(x) for all x ∈B.(b) Pick anyk ∈ {1,⋯,n}, and definea(k) = x− rek

,b(k) = x+ rek, whereek is thek-thunit vector inRn. Then, by definition ofB, we havea(k) ∈ B andb(k) ∈ B. And, sinceB is a convex set, we have[tb(k)+ (1− t)a(k)] ∈ B for all t ∈ I ≡ [0,1]. For t ∈ I , definef (t) =F(tb(k)+(1−t)a(k)), and note that this function is well defined, sinceF is definedon C (which contains the setB). Define A = (0,1), and verify that f is continuouslydifferentiable onA.

(c) Show thatf (1/2) = F(x), and f (1/2) ≥ f (t) for all t ∈A.(d) Use the Theorem (stated above) to obtainf ′(1/2) = (2r)DkF(x) = 0. Verify that

this proves the Corollary.

Solution.

(a) Sincex ∈C is a point of local maximum ofF , there is someβ > 0 such that for allx ∈C satisfyingd(x, x) < β, we haveF(x) ≤ F(x). And sinceC is open, there issomeγ > 0 such that for allx ∈ Rn satisfyingd(x, x) < γ, we havex ∈C. Now, letr = 1

2 min{β,γ}, so thatB≡ B(x,2r) = {x ∈Rn ∣ d(x, x) < 2r } ⊂C. Then for allx ∈ B,we have thatF(x) ≥ F(x).

(b) For anyk ∈ {1, . . . ,n}, define

a(k) = x− rek, b(k) = x+ rek


Now,a(k) ∈B andb(k) ∈B becaused(a(k), x) =d(b(k), x) =d(rek,0) = r <2r. Since

the open ballB is convex, we have thattb(k)+ (1− t)a(k) ∈ B for all t ∈ I = [0,1].Define f ∶ I →R by

f (t) = F(tb(k)+(1− t)a(k)) for all t ∈ I

Sincetb(k)+ (1− t)a(k) ∈ B ⊂C for all t ∈ I andF is defined onC, the function fis well-defined. LetA= (0,1). We showed in problem 1 of Problem Set 8 thatf iscontinuous on[0,1], so it follows thatf is continuous onA. SinceF is continuouslydifferentiable on the open setC, we can use the Chain Rule to differentiatef on A.For all t ∈A, we have

f ′(t) = n∑i=1[DiF(tb(k)+(1− t)a(k)) ∂

∂t(tbi(k)+(1− t)ai(k))]

=n∑

i=1[(bi(k)−ai(k)) DiF(tb(k)+(1− t)a(k))]

= (b(k)−a(k)) ∇F(tb(k)+(1− t)a(k))= (2rek) ∇F(x+(2t −1)rek)= 2rDkF(x+(2t −1)rek)

SinceF is continuously differentiable onC andtb(k)+(1− t)a(k) = x+(2t −1)rek ∈

C whenevert ∈ A, we have thatf ′(t) is continuous onA. So f is continuouslydifferentiable onA.

(c) Sincetb(k)+(1− t)a(k) = x+(2t −1)rek, we havef (12) = F(x). And sincetb(k)+(1− t)a(k) ∈B for all t ∈ I andF(x) ≥ F(x) for all x ∈B, we have by the definition of

f that f (12) ≥ f (t) for all t ∈ I , and thus for allt ∈A.

(d) Using the result in part (c), we have by the stated Theorem that f ′(12) = 0. And from

part (b), we have thatf ′(12) = 2rDkF(x). Sincer > 0, these two equations imply that

DkF(x) = 0. Becausek ∈ {1, . . . ,n} was chosen arbitrarily, we have∇F(x) = 0. Thisproves the Corollary.

Problem 37(Unconstrained Optimization: Sufficient Conditions for a Global Maximum).

SupposeA is an open convex set inRn, and f ∶A→R is twice continuously differentiableand quasi-concave onA. Suppose there is ¯x in A satisfying:

(i) ∇ f (x) = 0, and (ii)H f (x) is negative definite.Show that ¯x is the unique point of global maximum off onA.


Solution.

We will prove this by verifying a series of claims. First, we will show that the Hessianof f is negative definite in some neighborhood around ¯x. Second, we will use Taylor’sTheorem to show that ¯x is a point of strict local maximum off . Finally, we will use thequasi-concavity off onA to show that ¯x is the unique point of global maximum off onA.Claim 1: There is someδ > 0 such that for allx ∈ B(x,δ) = {x ∈Rn ∣ d(x, x) < δ}, we havethatx ∈A andH f (x) is negative definite.Proof: For eachr = 1, . . . ,n, let F r ∶A→ R denote therth leading principal minor of theHessian off . SinceF r is constructed by summing products of second derivatives off ,which are continuous onA, it follows that eachF r is continuous onA.

SinceH f (x) is negative definite, we know thatF1(x) =D11 f (x) < 0. By the continuityof F1, for ε1 = −F1(x) > 0 there is someδ1 > 0 such that wheneverx ∈ A andd(x, x) < δ1,we have∣F1(x)−F1(x)∣ < ε1, which implies thatF1(x) < 0.

We can constructδ2, . . . ,δn similarly. Then for eachr =1, . . . ,n we have that for allx∈A

satisfyingd(x, x) < δr , it follows thatF r(x) takes the appropriate sign (strictly negative foroddr and strictly positive for evenr).

SinceA is open, there is someγ > 0 such that for allx ∈ Rn satisfyingd(x, x) < γ, wehavex ∈A.

Now, defineδ =min{δ1, . . . ,δn

,γ} > 0. Then for allx ∈ Rn satisfyingd(x, x) < δ, wehave thatx ∈A andH f (x) is negative definite. This proves the claim.Claim 2: x is a point of strict local maximum off .Proof: Note that by Theorem 37, the conditions in the problemimply that x is a point oflocal maximum off . We want to use Claim 1 to argue that we can strengthen this resultto show that ¯x is a point of strict local maximum off .

Consider somex′ such thatx′ ≠ x andd(x′, x) < δ. By Claim 1, we have thatx′ ∈A, so fis well-defined atx′. We want to show thatf (x) > f (x′). SinceA is open and convex andf ∶A→R is twice continuously differentiable onA, we can apply Taylor’s Theorem, whichsays that there isθ ∈ (0,1) such that

f (x′)− f (x) = (x′− x)∇ f (x)+ 12[(x′− x)′H f (θx′+(1−θ)x)(x′− x)]

We are given that∇ f (x) = 0. Also, sinceB(x,δ) is a convex set andx′, x∈B(x,δ), we havethat θx′ + (1−θ)x ∈ B(x,δ), so it follows by Claim 1 thatH f (θx′ + (1−θ)x) is negativedefinite. Becausex′− x≠ 0, this negative definiteness means we can use the above equationto write

f (x′)− f (x) = 12[(x′− x)′H f (θx′+(1−θ)x)(x′− x)] < 0

This implies f (x′) < f (x), which shows that ¯x is a point of strict local maximum off .Claim 3: x is the unique point of global maximum off onA.


Proof: Seeking contradiction, suppose the claim does not hold. Then there is some ˜x ∈ Asuch that ˜x≠ x and f (x) ≥ f (x). Now, we can take someλ ∈ (0,1) sufficiently close to zerosuch thatx′′ = λx+ (1−λ)x ∈ B(x,δ). It follows by Claim 2 thatf (x) > f (x′′). But sincef (x) ≥ f (x) and f is quasi-concave onA, we have thatf (x′′) = f (λx+ (1−λ)x) ≥ f (x).This is a contradiction, so the claim holds.


References:This material is based onThe Elements of Real Analysisby R. Bartle (Chapter 7);

Mathematical Analysisby T. Apostol(Chapter 6); andMathematical Economicsby A.Takayama(Chapter 1).

Chapter 9

Constrained Optimization

9.1 Preliminaries

Let A be a subset ofRn, and f , g be real-valued functions onA. Define theconstraintset, C ≡ {x ∈ A ∶ g(x) = 0}. A point x∗ ∈C is a point of local maximum of f subject to theconstraint g(x) =0, if there isδ >0, such thatx∈C∩B(x∗,δ) implies f (x) ≤ f (x∗). A pointx∗ ∈C is a point of global maximum of f subject to the constraint g(x)=0, if x∗ solves theproblem:

Max f(x)sub ject tox∈C

9.2 Necessary Conditions for a Constrained Local Maxi-mum

The basic necessary condition for a constrained local maximum is provided by Lagrange’stheorem.

Theorem 40. (Lagrange)Let A⊂ Rnbe open, and f∶ A→ R, g ∶ A→ R be continuously differentiable functions

on A. Suppose x∗ is a point of local maximum of f subject to the constraint g(x) = 0.Suppose, further, that∇g(x∗) ≠ 0. Then there isλ∗ ∈R such that

[First-Order Condition] ∇ f (x∗) = λ∗∇g(x∗)Remark: There is an easy way of remembering the conclusion of the theorem. We write

L(x;λ) = f (x)−λg(x)141

CHAPTER 9. CONSTRAINED OPTIMIZATION 142

whereL ∶A×R→R. L is known as the “Lagrangian”, andλ as the “Lagrange multiplier”.Consider now the problem of finding the local maximum in anunconstrained maximiza-tion problemin whichL is the function to be maximized. The first-order conditions are

DiL(x,λ) = 0 f or i = 1, ...,n+1

This yieldsDi f (x) = λDig(x) i = 1, ...,n

andg(x) = 0

The firstn equations can be written as

∇ f (x) = λ∇g(x)The method described above is known as the “Lagrange multiplier method”.The Constraint Qualification

It is particularly important to check the condition∇g(x∗) ≠ 0, before applying the con-clusion of Lagrange’s theorem. This condition is known as the constraint qualification.Without this condition, the conclusion of Lagrange’s theorem would not be valid, as thefollowing example shows.Example

Let f ∶R2→R be given byf (x1, x2) = 2x1+3x2 for all (x1, x2) ∈R2; let g ∶R2

→R begiven byg(x1, x2) = x2

1+x22. Consider the constraint setC = {(x1, x2) ∈R2 ∶ g(x1,x2) = 0}.

The only element of this set is (0,0), so(x∗1, x∗2) = (0,0) is a point of local maximum offsubject to the constraintg(x) = 0.

The conclusion of Lagrange’s theorem does not hold here. For, if it did, there wouldexistλ∗ ∈R such that ∇ f (0,0) = λ∗∇g(0,0)But this means that (2,3) = (0,0)which is clearly a contradiction. The problem here is that∇g(x∗1, x∗2) =∇g(0,0) = (0,0),so the constraint qualification is violated.

Theorem 41. Let A ⊂ Rn be open, and f∶ A→ R, g ∶ A→ R be twice continuously dif-ferentiable functions on A. Suppose x∗ is a point of local maximum of f subject to theconstraint g(x) = 0. Suppose, further, that∇g(x∗) ≠ 0. Then there isλ∗ ∈R such that


[First-Order Condition} ∇ f (x∗) = λ∗∇g(x∗)[Second-Order Necessary Condition]yHL(x∗, λ∗)y≤ 0 for all y satisfying

y∇g(x∗) = 0

where L(x;λ∗) = f (x)−λ∗g(x) for all x ∈ A, and HL is the n×n Hessian matrix of L withrespect to(x1, ...,xn).9.3 The Arithmetic Mean-Geometric Mean Inequality

Consider the constrained maximization problem

MaxnΠi=1

xi

sub ject ton∑

i=1xi = n

and xi ≥ 0 i = 1, ...,n

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭(P)

DefineC = {x ∈ Rn+ ∶ n∑

i=1xi = 1). ThenC is a non-empty, closed and bounded set inRn.

DefineF ∶ Rn→R by F(x1, ....,xn) = n

Πi=1

xi. ThenF is a continuous function onRn. By

Weierstrass’ Theorem, there isx∗ ∈C, such thatF(x) ≤ F(x∗) for all x ∈C. That is,x∗

solves (P). Clearly,x∗i > 0 for all i. We can therefore conclude thatx∗ also solves thefollowing problem:

MaxnΠi=1

xi

sub ject ton∑

i=1xi = n

and xi > 0 i = 1, ...,n

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭(Q)

DefineA = Rn++; thenA is an open subset ofRn. Defineg ∶ Rn

++ → R by g(x1, ...,xn) =n∑

i=1xi −n, and f ∶Rn

++ →R by f (x1, ...,xn) = nΠi=1

xi. Thenx∗ solves(Q), sox∗ is a point of

local maximum of f subject to the constraintg(x) = 0. Also,∇g(x∗) = (1,1, ...,1) ≠ 0.So, by the Lagrange theorem, there isλ ∈R, such that

∇ f (x∗) = λ∇g(x∗)Definingy∗ ≡

nΠi=1

x∗i , we obtain

[(y∗/x∗1), ...,(y∗/x∗n)] = [λ, ....,λ]


Thusλ > 0, andx∗i = (y∗/λ) for all i, so thatx∗1 = x∗2 = ... = x∗n. Sincen∑

i=1x∗i = n, we getx∗i = 1

for all i. ThusnΠi=1

x∗i =1. To summarize, we have now demonstrated that if(x1, ...,xn) ∈Rn+,

andn∑

i=1xi = n, then

nΠi=1

xi ≤ 1.

Let a1, ...,an ben positive numbers. Defineα =n∑

i=1ai , andxi = [nai/α] for i = 1, ...,n.

Then(x1, ...,xn) ∈Rn+ and

n∑i=1

xi = n. So by our conclusion in the preceding paragraph, we

havenΠi=1

xi ≤ 1. This meansnΠi=1(nai/α) ≤ 1

so,

(na)n n

Πi=1

ai ≤ 1

and,

nΠi=1

ai ≤ (αn)n

=

⎡⎢⎢⎢⎢⎢⎢⎢⎣

n∑i=1

ai

n

⎤⎥⎥⎥⎥⎥⎥⎥⎦

n

This yields finally

( nΠi=1

ai)1/n ≤n∑

i=1ai

nwhich is the Arithmetic Mean - Geometric Mean inequality.

9.4 Sufficient Conditions for a Constrained Local Maxi-mum

Theorem 42. Let A⊂Rn be open, and f∶A→R, g ∶A→R be twice continuously differen-tiable functions on A. Suppose(x∗,λ∗) ∈C×R and

[First-Order Condition] ∇ f (x∗) = λ∗∇g(x∗)[Second-Order Sufficient Condition] yHL(x∗,λ∗)y<0 for all y≠0satisfying y∇g(x∗)=0


where L(x,λ∗) = f (x)−λ∗g(x) for all x ∈ A and HL is the n×n Hessian matrix of L withrespect to(x1, ....,xn). Then, x∗ is a point of local maximum of f subject to the constraintg(x) = 0.

There is a convenient method of checking thesecond-order sufficient conditionstatedin the above theorem, by checking the signs of the leading principal minors of the relevant“bordered” matrix. This method is stated in the following Proposition.

Proposition 5. Let B be an n×n symmetric matrix, and a be an n-vector with a1 ≠ 0.Define the(n+1)×(n+1)matrix S by

S= [ 0 aa B

]and let∣S(k)∣ be the(k+1)th leading principal minor of S for k= 1, ...,n. Then the follow-ing two statements are equivalent∶

(i) yBy< 0 for all y ≠ 0 such that ya= 0(ii) (−1)k ∣S(k)∣ > 0 for k= 1, ...,n.

9.5 Sufficient Conditions for a Global Maximum

Theorem 43. Let A⊂Rn be an open convex set, and f∶A→R, g ∶A→R be continuouslydifferentiable functions on A. Suppose(x∗,λ∗) ∈C×R satisfies[First-Order Condition] ∇ f (x∗) = λ∗∇g(x∗)If L(x,λ∗) ≡ f (x)−λ∗g(x) is concave in x on A, then x∗ is a point of global maximum off subject to the constraint g(x) = 0.

To see this, letx ∈C. Then,

L(x,λ∗)−L(x∗,λ∗) ≤ (x−x∗)[∇ f (x∗)−λ∗∇g(x∗)]by concavity ofL in x onA. Using the first-order condition, we get

f (x)−λ∗g(x) = L(x,λ∗) ≤ L(x∗,λ∗) = f (x∗)−λ∗g(x∗).Sincex ∈C, andx∗ ∈C, we haveg(x) = g(x∗) = 0. Thus,f (x) ≤ f (x∗), and sox∗ is a pointof global maximum off subject to the constraintg(x) = 0.Example: Let f ∶ R2

→ R be given by f (x,y) = (1− x2− y2); g ∶ R2→ R be given by

g(x,y) = x+4y−2.


To apply the sufficient conditions, set up the Lagrangian

L(x,y,λ) = (1−x2−y2)−λ[x+4y−2]The first-order conditions yield

−(2x∗, 2y∗) = λ∗[1,4] (9.1)

Also, to satisfy the constraint,x∗+4y∗ = 2 (9.2)

Thus, from (9.1),x∗ = −(λ∗/2), y∗ = −2λ∗. Using this in (9.2),

−(λ∗/2)−8λ∗ = 2 (9.3)

which yieldsλ∗ = −(4/17), x∗ = (2/17), y∗ = (8/17).To check the second-order sufficient condition, we write therelevant bordered matrix

S= [ 0 ∇g(x∗,y∗)∇g(x∗,y∗) HL(x∗,y∗,λ∗) ] =⎡⎢⎢⎢⎢⎢⎣

0 1 41 −2 04 0 −2

⎤⎥⎥⎥⎥⎥⎦∣S(1)∣ = ∣ 0 1

1 −2 ∣ = −1< 0

∣S(2)∣ = (−1)∣ 1 04 −2 ∣+4∣ 1 −2

4 0∣

= 2+32= 34> 0

Thus(−1)k ∣S(k)∣ > 0 fork = 1,2; by Proposition 5,zHL(x∗,y∗;λ∗)z< 0 for all z≠ 0 suchthat z∇g(x∗,y∗) = 0. Thus, the second-order condition of Theorem 42 is satisfied, andso (x∗,y∗) = (2/17,8/17) is a point of local maximum off subject to the constraintsg(x,y) = 0.

Notice that the Hessian ofL with respect to(x,y) is

HL = [ −2 00 −2 ]

which is negative definite. SinceA=R2 is a convex set,L(x,y,λ∗) is concave in(x,y) onA. Hence, using Theorem 43,(x∗,y∗) = (2/17,8/17) is a point of global maximum offsubject to the constraintg(x,y) = 0.



Problem 38 (Constrained Optimization: Proof of the Lagrange Theorem, using the Im-plicit Function Theorem).

Let A be an open subset ofRn, and letf andg be continuously differentiable functionsfrom A to R. DefineC= {x ∈A ∶ g(x) = 0}. Assume that there is ¯x ∈C such that:

f (x) ≤ f (x) for all x ∈Cand Dng(x) ≠ 0

} (1)

(a) Use the implicit function theorem to show that there is anopen setX ⊂Rn−1 whichcontains(x1,⋯, xn−1) and an open setY ⊂ R which contains ¯xn, and a unique functionh ∶X→Y such that(x1,⋯,xn−1,h(x1,⋯,xn−1)) ∈A for all (x1,⋯,xn−1) ∈X, and:

(i) g(x1, ...,xn−1,h(x1, ...,xn−1)) = 0 for all (x1, ...,xn−1) ∈X(ii) h(x1, ..., xn−1) = xn} (2)

Further,h is continuously differentiable onX.

(b) Using (2), we have(x1, ...,xn−1,h(x1, ...,xn−1)) ∈C for all (x1, ...,xn−1) ∈ X, and soby (1) we have, for all(x1, ...,xn−1) ∈X ∶

f (x1, ...,xn−1,h(x1, ...,xn−1)) ≤ f (x1, ..., xn−1, h(x1, ..., xn−1)) (3)

Use (2) and (3), and the Corollary in problem 1 above, to prove that if we defineλ =Dn f (x)/Dng(x), then: ∇ f (x) = λ∇g(x) (4)

Solution.

(a) We want to show the existence of a Lagrange multiplier, giventhe knowledge thatx ∈C solves the constrained optimization problem. We are told that

f (x) ≤ f (x) for all x ∈C andDng(x) ≠ 0 (1)

Treatingxn as the variable andx1, . . . ,xn−1 as parameters, we want to apply the Im-plicit Function Theorem tog at the point ¯x= (x1, . . . , xn−1, xn) ∈A. We need to checkthe following conditions:

● The functiong is defined and continuously differentiable on the open setA ⊂Rn.


● We are told that ¯x ∈C, which impliesg(x) = 0.

● We are given thatDng(x) ≠ 0. Sincexn is the only variable, this is the onlyderivative condition we need to check.

By the Implicit Function Theorem, then, there is an open setX ⊂ Rn−1 containing(x1, . . . , xn−1), an open setY ⊂R containing ¯xn, and a unique functionh∶X→Y suchthat(x1, . . . ,xn−1,h(x1, . . . ,xn−1)) ∈A for all (x1, . . . ,xn−1) ∈X and

(i) g(x1, . . . ,xn−1,h(x1, . . . ,xn−1)) = 0 for all (x1, . . . ,xn−1) ∈X(ii) xn = h(x1, . . . , xn−1) } (2)

Further,h is continuously differentiable onX.

(b) By result (2)(i) above, we have that(x1, . . . ,xn−1,h(x1, . . . ,xn−1)) ∈C whenever(x1, . . . ,xn−1) ∈X. Then sincef (x) ≤ f (x) for all x ∈C, we have that for all(x1, . . . ,xn−1) ∈X,

f (x1, . . . ,xn−1,h(x1, . . . ,xn−1)) ≤ f (x1, . . . , xn−1,h(x1, . . . , xn−1)) (3)

SinceDng(x) ≠ 0 we can defineλ = Dn f (x)Dng(x) . We want to show that

∇ f (x) = λ∇g(x) (4)

That is, we want to show thatDi f (x) = λDig(x) for all i = 1, . . . ,n−1,n. Note thatthe last of thesen equations,Dn f (x) = λDng(x), holds by the definition ofλ. Now,define the functionF ∶X→R by

F(x1, . . . ,xn−1) = f (x1, . . . ,xn−1,h(x1, . . . ,xn−1)) for all (x1, . . . ,xn−1) ∈X

Since(x1, . . . ,xn−1,h(x1, . . . ,xn−1)) ∈C ⊂ A whenever(x1, . . . ,xn−1) ∈ X and sincefis defined onA, the functionF is well-defined. We know thath is continuously dif-ferentiable onX. Also, we found in part (a) that whenever(x1, . . . ,xn−1) ∈X we have(x1, . . . ,xn−1,h(x1, . . . ,xn−1)) ∈A, and we know thatf is continuously differentiableonA. So we can use the Chain Rule to find the derivative ofF on the open setX. Wecan also use the Chain Rule to differentiate result (2)(i) above. For alli = 1, . . . ,n−1and all(x1, . . . ,xn−1) ∈X, we have

DiF(x1, . . . ,xn−1) =Di f (x1, . . . ,xn−1,h(x1, . . . ,xn−1))+Dn f (x1, . . . ,xn−1,h(x1, . . . ,xn−1))Dih(x1, . . . ,xn−1) (5)

0=Dig(x1, . . . ,xn−1,h(x1, . . . ,xn−1))+Dng(x1, . . . ,xn−1,h(x1, . . . ,xn−1))Dih(x1, . . . ,xn−1) (6)


We have by (3) that(x1, . . . , xn−1) is a point of local maximum ofF . Now, because(x1, . . . , xn−1) ∈X, X is open, andF is continuously differentiable onX, we have bythe Corollary in problem 1 that∇F(x1, . . . , xn−1) = 0. That is,DiF(x1, . . . , xn−1) = 0for all i = 1, . . . ,n−1. We also know from (2)(ii) that ¯xn = h(x1, . . . , xn−1). Evaluating(5) and (6) at(x1, . . . , xn−1), then, for alli = 1, . . . ,n−1 we have

Di f (x1, . . . , xn−1, xn)+Dn f (x1, . . . , xn−1, xn)Dih(x1, . . . , xn−1) = 0 (7)

Dig(x1, . . . , xn−1, xn)+Dng(x1, . . . , xn−1, xn)Dih(x1, . . . , xn−1) = 0 (8)

Now, sinceDng(x) ≠ 0 we can solve (8) forDih(x1, . . . , xn−1) and plug this expres-sion into (7) to obtain that for alli = 1, . . . ,n−1,

Di f (x)−Dn f (x)Dig(x)Dng(x) = 0 (9)

Using the fact thatλ = Dn f (x)Dng(x) , we have from (9) that for alli = 1, . . . ,n−1,

Di f (x) = λDig(x) (10)

SinceDn f (x) = λDng(x) holds by the definition ofλ, we have by (10) that (4) holds,which is what we wanted to show.

Problem 39(Constrained Optimization: Sufficient Conditions for a Global Maximum).

SupposeA is an open convex set inRn, f ∶A→R is continuously differentiable and concave

onA, andg ∶A→R is a linear function onA.Suppose there is ¯x in A andλ ∈R which satisfy the following conditions:

(i) g(x) = 0(ii)∇ f (x) = λ∇g(x) }(FOC)Show that ¯x solves the problem:

Max f(x)sub ject to g(x) = 0and x∈A

⎫⎪⎪⎪⎬⎪⎪⎪⎭(P)Solution.


Given theλ in the hypothesis of the theorem, define the functionL∶A→R by

L(x) = f (x)−λg(x) for all x ∈A

Becausef is concave onA, g is linear onA, andλ is fixed, we have thatL is concaveon A. Moreover,L is continuously differentiable onA becausef andg are continuouslydifferentiable onA. In addition, we are given thatA is open and convex. So we can applyTheorem 27 to see that for allx ∈A satisfying the constraintg(x) = 0, we have

L(x)−L(x) ≤ (x− x)∇L(x)f (x)− f (x)−λ(g(x)−g(x)) ≤ (x− x)(∇ f (x)−λ∇g(x))

f (x)− f (x) ≤ 0

This shows that ¯x solves the problem(P).Problem 40(Sufficient Conditions for Constrained Maximization: Application).

We want to solve the problem:

Max ∏ni=1 xi

sub ject to ∑ni=1xi = n

and x∈Rn+

⎫⎪⎪⎪⎬⎪⎪⎪⎭(Q)by using the sufficient conditions for constrained maximization, developed in problem 4above.

Instead of solving(Q) directly, we look at the following problem:

Max ∑ni=1 ln xi

sub ject to ∑ni=1xi −n= 0

and x∈Rn++

⎫⎪⎪⎪⎬⎪⎪⎪⎭(R)(a) DefineA=Rn

++, f ∶A→R by f (x) =∑ni=1 ln xi for all x ∈A, andg ∶A→R by g(x) =

∑ni=1xi −n for all x ∈ A. Verify thatA, f andg satisfy all the hypotheses stated in problem

4.(b) Show that ¯x= (1,1, ...,1) andλ = 1 satisfy the(FOC) of problem 4. Use the result

of problem 4 to infer that ¯x= (1,1, ...,1) solves problem(R).(c) Verify, using (b), that ¯x= (1,1, ...,1) solves:

Max ∏ni=1 xi

sub ject to ∑ni=1xi −n= 0

and x∈Rn++

⎫⎪⎪⎪⎬⎪⎪⎪⎭(R′)

(d) Conclude, using (c), that ¯x= (1,1, ...,1) solves(Q).


Solution.

(a) DefineA=Rn++ and note thatA is an open and convex set inRn. Now, definef ∶A→R

andg∶A→R by

f (x) = n∑i=1

lnxi for all x ∈A, g(x) = n∑i=1

xi −n for all x ∈A

The function f can be expressed as the sum of functions that are continuously dif-ferentiable and concave onA, so f is also continuously differentiable and concaveon A. Also, g is linear onA. ThereforeA, f , andg satisfy all the hypotheses statedin problem 4.

(b) It is straightforward to check that when ¯x= (1, . . . ,1) ∈A andλ = 1, we haveg(x) = 0and∇ f (x) = λ∇g(x). Therefore by the sufficiency result in problem 4, ¯x solves theproblem(R).

(c) Define the functionh∶A→R by

h(x) = n∏i=1

xi for all x ∈A

Then for allx∈A, we have lnh(x)= f (x), which is equivalent toh(x)=ef (x). Seekingcontradiction, suppose that ¯x solves(R) but does not solve(R′). Then there isx′ ∈A,x′ ≠ x satisfyingg(x′)=0 andh(x′)>h(x), which is equivalent toef (x′) >ef (x). Sincethe exponential function is strictly increasing onR, this implies thatf (x′) > f (x).But that contradicts ¯x solving(R), so it must be that ¯x solves both(R) and(R′).

(d) We have from part (c) thath(x) ≥ h(x) for all x ∈ A=Rn++ satisfyingg(x) = 0. Con-

sider the set

S=Rn+∖Rn

++ = {x ∈Rn+ ∣ xi = 0 for somei ∈ {1, . . . ,n}}

For allx ∈S, we have thath(x) = 0< 1= h(x). Therefore ¯x solves(Q).


ReferencesThis material is based onThe Elements of Real Analysisby R. Bartle (Chapter 7);

Mathematical Analysisby T. Apostol(Chapter 6); andMathematical Ecoomicsby A.Takayama(Chapter 1).

The theory of maximum of a function of several variables is a useful tool to provemany important inequalities. For more on this, seeInequalitiesby G.H. Hardy, J.E. Lit-tlewood and G. Polya(Chapter 4). A proof of the important Proposition 5 can be found inH.B. Mann: “Quadratic Forms with Linear Constraints”,American Mathematical Monthly(1943), 430-433, and also in G. Debreu “Definite and Semidefinite Quadratic Forms”,Econometrica(1952), 295-300.

Part IV

Modern Optimization Theory

153

Chapter 10

Concave Programming

10.1 Preliminaries

We will present below the elements of “modern optimization theory” as formulated byKuhn and Tucker, and a number of authors who have followed their general approach. Asin our exposition of “classical optimization theory”, we will concentrate on characterizingpoints ofmaximumof a function of several variables (subject to certain constraints). Thetheory which characterizes points ofminimumof a function of several variables (subjectto certain constraints) can be obtained analogously.

Modern constrained maximization theory is concerned with the following problem:

Max f(x)Sub ject to gj(x) ≥ 0 f or j = 1, ...,m

and x∈X

⎫⎪⎪⎪⎬⎪⎪⎪⎭ (P)

whereX is a non-empty subset ofRn, and f , g j( j = 1, ...,m) are functions fromX to R.We define theconstraint set, C as follows:

C= {x ∈X ∶ g(x) ≥ 0}where, as usual,g(x) = [g1(x), ...,gm(x)].

An element ˆx ∈ X is a point of constrained global maximumif x solves the problem(P). A pair (x, λ) ∈ (X×Rm

+) is asaddle pointif

φ(x, λ) ≤ φ(x, λ) ≤ φ(x,λ)for all x ∈X and allλ ∈Rm

+ , whereφ(x,λ) = f (x)+λg(x) for (x,λ) ∈ X×Rm+ .

154

CHAPTER 10. CONCAVE PROGRAMMING 155

10.2 Constrained Global Maxima and Saddle Points

A major part of modern optimization theory is concerned withestablishing (under suitableconditions) an equivalence result between a point of constrained global maximum and asaddle point. We explore this theory in what follows.

Theorem 44. If (x, λ) ∈ (X ×Rm+) is a saddle point, then (i)λg(x) = 0, (ii) g(x) ≥ 0, and

(iii) x is a point of constrained global maximum.

Proof. Since(x, λ) is a saddle point, we have for allλ ∈Rm+ ,

f (x)+ λg(x) ≤ f (x)+λg(x)That is, we have

λg(x) ≤ λg(x) f or all λ ∈Rm+ (10.1)

Choosingλ = 0 in (10.1), we getλg(x) ≤ 0. Choosingλ = 2λ in (10.1), we getλg(x) ≥ 0.Thusλg(x) = 0, which proves (i).

Using (i) in (10.1), we get

0≤ λg(x) f orall λ ∈Rm+ (10.2)

Choosingλ in turn to be themunit vectors inRm in (10.2), we get

0≤ g(x) (10.3)

which proves (ii). Thus ˆx is in the constraint set,C. Now, letx be an arbitrary element ofC. Since(x, λ) is a saddle point, we have

f (x)+ λg(x) ≥ f (x)+ λg(x) (10.4)

Using (i), and the fact thatλ ∈Rm+ in (10.4), we obtain

f (x) ≥ f (x)Thus,x solves (P), proving (iii).//

A converse of Theorem 44 can be proved ifX is a convex set,f , g j( j = 1, ...,m) areconcave functions onX, and a condition on the constraints generally known as “Slater’scondition” is satisfied. [Notice that none of these conditions are needed for the validity ofTheorem 44].

Given the problem (P), we will say thatSlater’s conditionholds if there is ¯x ∈ X, suchthatg j(x) > 0 for j = 1, ...,m.


Theorem 45. (Kuhn-Tucker) Supposex ∈ X is a point of constrained global maximum.If X is a convex set, f,g j( j = 1, ...,m) are concave functions on X, and Slater’s conditionholds, then there isλ ∈Rm

+ such that (i)λg(x) = 0, and (ii) (x, λ) is a saddle point.

Proof. Define the setsA andB as follows:A= {(α,β) ∈Rm+1 ∶ f (x)− f (x) ≥ α andg(x) ≥ β for somex ∈X}B= {(α,β) ∈Rm+1 ∶ α > 0 andβ >> 0}

ThenB is clearly a non-empty, convex set. AndA is a non-empty, convex set, sinceXis convex, andf , g j( j = 1, ...,m) are concave functions. Sincex solves(P), A andB aredisjoint. Using the Minkowski Separation theorem, we have(µ,ν) ∈Rm+1

, (µ,ν) ≠ 0, andθ ∈R, such that

µα+νβ ≤ θ f orall (α,β) ∈A (10.5)

µα+νβ ≥ θ f orall (α,β) ∈B (10.6)

Using (10.5),θ ≥ 0, while using (10.6),θ ≤ 0. Thusθ = 0. Also, using (10.6),µ≥ 0 andν ≥ 0. Summarizing, we have(µ,ν) ∈Rm+1

+ ,(µ,ν) ≠ 0, such that

µ[ f (x)− f (x)]+νg(x) ≤ 0 (10.7)

for all x ∈X.We claim, now, thatµ≠ 0. For if µ= 0, thenν ≠ 0, and (10.7) yields

νg(x) ≤ 0 f orall x ∈X (10.8)

By Slater’s Condition, there isx ∈ X with g(x) >> 0. Sinceν ≥ 0 andν ≠ 0, soνg(x) > 0,which contradicts (10.8). Thus,µ≠ 0; that is,µ> 0. Defineλ = (ν/µ). Thenλ ∈Rm

+ and(10.7) yields

f (x)+ λg(x) ≤ f (x) f orall x ∈X (10.9)

Puttingx= x in (10.9), we getλg(x) ≤ 0 (10.10)

Also g(x) ≥ 0, andλ ∈Rm+ implies

λg(x) ≥ 0 (10.11)

Clearly (10.10) and (10.11) implyλg(x) = 0, which proves (i).Using (i) in (10.9), we get

f (x)+ λg(x) ≤ f (x)+ λg(x) f orall x ∈X (10.12)

If λ ∈Rm+ , thenλg(x) ≥ 0, sinceg(x) ≥ 0. Thus, using (i) again, we have

f (x)+ λg(x) ≤ f (x)+λg(x) f orall λ ∈Rm+ (10.13)


Combining (10.12) and (10.13), one can conclude that(x, λ) is a saddle point, whichestablishes (ii).

The following examples demonstrate why the assumptions of Theorem 45 are neededfor the conclusion to be valid.Example: Let X = R+; let f ∶ X → R be given by f (x) = x, andg ∶ X → R be given byg(x) = −x2. ThenX is convex, f andg are concave. But Slater’s Condition is clearlyviolated. It is easily checked that ˆx = 0 is a point of constrained global maximum. Butthere is noλ ∈R+ such that(x, λ) is a saddle point. For if there were such aλ, then

x− λx2 ≤ x2− λx2 = 0

for all x∈X. But by choosingx>0 andx sufficiently close to zero, this inequality is clearlyviolated.Example: Let X = R+; let f ∶ X → R be given byf (x) = x2, andg ∶ X → R be given byg(x) = 1−x. Here,X is convex,g is concave, and Slater’s condition is satisfied with (forinstance)x= (1/2). Buf f is not concave onX. It is easily checked that ˆx= 1 is a pointof constrained global maximum. But there is noλ ∈R+ such that(x, λ) is a saddle point.For if there were such aλ, then

x2+ λ(1−x) ≤ x2+ λ(1− x) = 1

for all x ∈ X. But by choosingx > 0 andx sufficiently large, this inequality is clearlyviolated.Example: Let X = R+; let f ∶ X → R be given by f (x) = x, andg ∶ X → R be given byg(x) = 1−x1/2. ThenX is convex, f is concave, and Slater’s condition is satisfied with(for instance)x = (1/4). But g is not concave onX. It can be checked that ˆx= 1 is a pointof constrained global maximum. But there is noλ ∈R+, such that(x, λ) is a saddle point.For, if there were such aλ, then

x+ λ(1−x1/2) ≤ x+ λ(1− x1/2) = 1

for all x ∈ X. But, by choosingx > 0 andx sufficiently large, this inequality is clearlyviolated.


10.3 The Kuhn-Tucker Conditions and Saddle Points

Let X be an open set inRn, and f , g j( j = 1, ...,m) be continuously differentiable onX. Apair (x, λ) in X×Rm

+ satisfies theKuhn-Tucker conditionsif

(i) Di f (x)+ m∑j=1

λ jDig j(x) = 0 i = 1, ...,n

(ii) g(x) ≥ 0 and λg(x) = 0

A part of modern optimization theory is concerned with establishing the equivalence (un-der some suitable conditions) between a saddle point and a point where the Kuhn-Tuckerconditions are satisfied. We examine this theory in what follows.

Theorem 46. Let X be an open set inRn, and f, g j( j = 1, ...,m) be continuously differen-tiable on X. Suppose a pair(x, λ) ∈ X×Rm

+ satisfies the Kuhn-Tucker conditions. If X isconvex and f , gj[ j = 1, ...,m] are concave on X, then (i)(x, λ) is a saddle point, and (ii)xis a point of constrained global maximum.

Proof. Define, as usual,φ(x,λ) = f (x)+λg(x) for (x,λ) ∈ X ×Rm+ . Given λ, φ(x, λ) is

concave inx, since f andg j( j = 1, ...,n) are concave inx. Thus forx ∈X we have

φ(x, λ)−φ(x, λ) ≤ (x− x)∇φ(x, λ)Using the Kuhn-Tucker conditions, we have∇φ(x, λ) = 0, so

φ(x, λ) ≤ φ(x, λ) f orall x ∈X (10.14)

Also, by the Kuhn-Tucker conditions,g(x) ≥ 0 andλg(x) = 0. So, for allλ ∈Rm+ , φ(x,λ) =

f (x)+λg(x) ≥ f (x) = f (x)+ λg(x) = φ(x, λ). Thus, we have

φ(x, λ) ≤ φ(x,λ) f orall λ ∈Rm+ (10.15)

Using (10.14) and (10.15),(x, λ) is a saddle point, which proves (i). Using Theorem 44,x solves(P) which proves (ii).

Theorem 47. Let X be an open set inRn, and f , gj( j = 1, ...,m) be continuously differ-entiable on X. Suppose a pair(x, λ) ∈ X×Rm

+ is a saddle point. Then(x, λ) satisfies theKuhn-Tucker conditions.


Proof. Since(x, λ) is a saddle point, we have

φ(x, λ) ≤ φ(x, λ) f orall x ∈X

Thus, givenλ, φ(x, λ) attains a maximum at ˆx ∈X. SinceX is open, andφ is continuouslydifferentiable onX, we have ∇φ(x, λ) = 0

Thus, Di f (x)+ m∑j=1

λ j Dig j(x) = 0 for i = 1, ...,n. Also, by Theorem 44, we know that

g(x) ≥ 0 andλg(x) = 0. Thus(x, λ) satisfies the Kuhn-Tucker conditions.

10.4 The Kuhn-Tucker Conditions and Constrained Lo-cal Maxima

An element ˆx ∈ X is apoint of constrained local maximumif x ∈C and there isδ > 0 suchthat for allx ∈B(x,δ)∩C, f (x) ≥ f (x).

Let X be an open set inRn, and f , g j( j = 1, ...,m) be continuously differentiable onX. We now establish the useful result (corresponding to the classical Lagrange theorem)that if x ∈X is a point of constrained local maximum then under suitable conditions, thereis λ ∈Rm

+ such that(x, λ) satisfies the Kuhn-Tucker conditions. The important conditionneeded for the development of this theory is called a “constraint qualification” (just as it isin classical theory). While there are several versions of this condition, the following one,due to Arrow, Hurwicz and Uzawa, appears to be the most useful.

Let x be a point in the constraint setC. Let E(x) ⊂ {1, ...,m} be the set of indices,j,for which the constraints arebindingat x; that isg j(x) = 0 for j ∈ E(x). Thenx satisfiestheArrow-Hurwicz-Uzawa (AHU) constraint qualificationif at least one of the followingthree conditions is satisfied:

(a) E(x) is empty(b) X is a convex set andg j is a convex function for eachj ∈E(x)(c) X is a convex set and there ish ∈Rn such thath∇g j(x) > 0 for all j ∈E(x).

Theorem 48. (Arrow-Hurwicz-Uzawa) Let X be an open set inRn, and f , gj( j = 1, ...,m)be continuously differentiable on X. Supposex ∈ X is a point of constrained local maxi-mum. Suppose, further, thatx satisfies the Arrow-Hurwicz-Uzawa constraint qualificationcondition, then there isλ ∈Rm

+ such that(x, λ) satisfies the Kuhn-Tucker conditions.

Proof. If condition (a) of the AHU constraint qualification is satisfied, theng j(x) > 0 forj = 1, ...,m. DefineY = {x ∈ X ∶ g j(x) > 0 for j = 1, ...,m}. Thenx ∈Y is a point of local


maximum of f on the open setY. So, by the theory of unconstrained maximum,

∇ f (x) = 0

Defineλ j = 0 for j = 1, ...,m. Then,(x, λ) satisfies the Kuhn-Tucker conditions.If condition (a) is not satisfied, letE ≡E(x) be the set of indices,j, for whichg j(x) =0.

(This is the set of effective constraints at the point of constrained local maximum). Sincex is a point of constrained local maximum, there isδ > 0, such thatx ∈ B(x,δ)∩C impliesf (x) ≤ f (x). Let z∈Rn be an arbitrary vector satisfyingz∇g j(x) ≥ 0 for j ∈ E. We willnow show that

z∇ f (x) ≤ 0 (10.16)

if either condition (b) or condition (c) of the AHU constraint qualification is satisfied.First, suppose condition (b) is satisfied. By definingh = tz with t > t > 0 andt suf-

ficiently close to zero, we haveh ∈ Rn, (x+h) ∈ B(x,δ), g j(x+h) > 0 for j ∈∼ E, andh∇g j(x) ≥ 0 for j ∈ E. By convexity ofX and ofg j for j ∈ E, we haveg j(x+h)−g j(x) ≥h∇g j(x) ≥ 0 for j ∈E. Thus,g j(x+h) ≥ g j(x) ≥ 0 for j ∈E. So(x+h) ∈B(x,δ)∩C. Sincex is a point of constrained local maximum,f (x+h) ≤ f (x). Thus, by applying the MeanValue theorem, we get 0≥ f (x+h)− f (x) = tz∇ f (ξ) whereξ is a convex combination of ˆxand(x+ tz). Thus,z∇ f (ξ) ≤ 0. Lettingt → 0, we getξ→ x, and∇ f (ξ)→∇ f (x). So, weobtainz∇ f (x) ≤ 0, which is (10.16).

Next, suppose condition (c) is satisfied. Definey = λh+z, whereλ ∈ (0,1), andh isgiven by condition (c). Then, we havey∇g j(x) > 0 for all j ∈ E, sincez∇g j(x) ≥ 0 andh∇g j(x) > 0 for eachj ∈E.Definingb= ty with t > t > 0,andt sufficiently close to zero, wehaveb∈Rn

,(x+b) ∈B(x,δ), g j(x+b) > 0 for j ∈∼E, and for allθ ∈ [0,1], b∇g j(x+θb) > 0for j ∈ E. Now, for j ∈ E, by the Mean-Value theorem,g j(x+b)−g j(x) = b∇g j(x+θ jb)for someθ j ∈ [0,1]. So, for j ∈ E, g j(x+b)−g j(x) > 0,and consequentlyg j(x+b) > 0.Thus, (x+b) ∈ B(x,δ)∩C. Since x is a point of constrained local maximum, we havef (x+b) ≤ f (x). Now, following the argument used in the previous paragraph,we gety∇ f (x) ≤ 0. Thus, we have shown that for everyλ ∈ (0,1), y = λh+z satisfiesy∇ f (x) ≤0;that is,z∇ f (x)+λh∇ f (x) ≤ 0. Letting λ→ 0,we obtainz∇ f (x) ≤ 0, as we had claimed.

We have now established that ifz∈Rn is an arbitrary vector which satisfiesz∇g j(x) ≥0for j ∈ E, thenz[−∇ f (x)] ≥ 0. By the Farkas Lemma [see Chapter 7 on “Convex Analy-sis”], there existλ j ≥ 0 for j ∈E, such that fori = 1, ...,n

−Di f (x) =∑ λ jDig j(x)Defineλ j = 0 for j ∈∼E. Thenλ ∈Rm

+ , and fori = 1, ...,n,

Di f (x)+ m∑j=1

λ jDig j(x) = 0


Sincex ∈C, we getg(x) ≥ 0. Sinceg j(x) = 0 for j ∈ E, andλ j = 0 for j ∈∼ E, we obtainλg(x) = 0. Thus,(x, λ) satisfies the Kuhn-Tucker conditions.

Corollary 6. Suppose X is an open, convex set inRn, and f,g j( j = 1, ...,m) are con-tinuously differentiable on X. Supposex ∈ X is a point of constrained local maximum.Suppose, further, that at least one of the following two conditions is satisfied:

(i) g j is convex for all j= 1, ...,m(ii) g j is concave for all j= 1, ...,m, and Slater’s condition holds.

Then, there isλ ∈Rm+ such that(x, λ) satisfies the Kuhn-Tucker conditions.

Proof. Let E ≡ E(x) be the set of indices for whichg j(x) = 0. If E(x) is empty, thencondition (a) of the AHU constraint qualification is satisfied and the result follows fromTheorem 48. IfE is non-empty, and condition (i) is satisfied, then clearly condition (b) ofthe AHU constraint qualification is satisfied and, again, theresult follows from Theorem48.

SupposeE is non-empty, and condition (ii) is satisfied. Then, there isx ∈C, suchthatg j(x) > 0 for j = 1, ...,m. Thus, for j ∈ E, we haveg j(x)−g j(x) ≤ (x− x)∇g j(x), byconcavity ofg j . Sinceg j(x) = 0 andg j(x) > 0 for eachj ∈E, we obtain(x− x)∇g j(x) > 0for j ∈E. Definingh= (x− x) ∈Rn, we haveh∇g j(x) > 0 for all j ∈E, and so condition (c)of the AHU constraint qualification is satisfied, and the result follows from Theorem 48.

Corollary 7. Suppose c∈ Rn, b ∈ Rk and A is a k×n matrix. Consider the followingmaximization problem:

Max cxSub ject to Ax≤ b

and x∈Rn+

⎫⎪⎪⎪⎬⎪⎪⎪⎭ (P)Supposex solves(P), then there isµ∈Rk

+ such that for all x∈Rn+, and all µ∈Rk

+(i) cx+ µ(b−Ax) ≤ cx+ µ(b−Ax) ≤ cx+µ(b−Ax)(ii) µ solves the following minimization problem:

Min µbSub ject to µA≥ cand µ∈Rk

+

⎫⎪⎪⎪⎬⎪⎪⎪⎭ (Q)Proof. DefineX =Rn; thenX is an open, convex set inRn. Define f ∶Rn

→R by f (x)= cx;defineg j ∶Rn

→R by g j(x) = x j for j = 1, ...,n; g j(x) = (b−Ax)( j−n) for j = n+1, ...,n+k.Definem= n+k. Then f , g j( j = 1, ...,m) are continuously differentiable onX. Clearlyg j


is convex for allj =1, ...,m. Thus, there isλ ∈Rm+ such that(x, λ) satisfies the Kuhn-Tucker

conditions, by applying Corollary 6. We can writeλ = (ν, µ) whereν ∈ Rn+ andµ= Rk

+.Then, we have fori = 1, ...,n

ci + νi − k∑j=1

µja ji = 0

That is,c+ ν− µA= 0 (10.17)

andx≥ 0, (b−Ax) ≥ 0, ν x+ µ(b−Ax) = 0. Clearly, then,

ν x= 0andµ(b−Ax) = 0 (10.18)

Using (10.17), we get for allx ∈ Rn+, cx+ µ(b−Ax) = cx+ νx− µAx+ µb− νx ≤ µb. Also,

using (10.17) and (10.18), we get

cx+ µ(b−Ax) = cx+ ν x− µAx+ µb− ν x= µb (10.19)

Thus, for allx ∈Rn+

cx+ µ(b−Ax) ≤ cx+ µ(b−Ax) (10.20)

For all µ ∈Rk+, we havecx+µ(b−Ax) ≥ cx, since(b−Ax) ≥ 0. Alsocx+ µ(b−Ax) = cx,

sinceµ(b−Ax) = 0. Thus, for allµ∈Rk+,

cx+µ(b−Ax) ≥ cx+ µ(b−Ax (10.21)

Combining (10.20) and (10.21) establishes the result (i).To prove (ii), we can proceed as follows. We get from (10.18),(10.19) that

cx= µb (10.22)

Using this in (10.20), we get for allx∈Rn+, (c−µA)x+µb≤ µb+µ(b−Ax) and using (10.18)

in the above inequality (c− µA)x≤ 0 f orall x ∈Rn+ (10.23)

Choosingx in turn to be then unit vectors inRn, we get from (10.23),

µA≥ c (10.24)

Sinceµ∈Rk+, µ is in the constraint set of problem(Q).

Now, consider an arbitraryµ ∈ Rk+ such thatµA≥ c. Then(c−µA)x ≤ 0 since

x ∈Rm+ , and using this in (10.21) yields

µb≥ cx+ µ(b−Ax) (10.25)

Using (10.18) and (10.22) in (10.25) implies thatµb≥ µb. This proves that ˆµ solves(Q).


10.5 Constrained Local and Global Maxima

It is clear that if x is a point of constrained global maximum, then ˆx is also a point ofconstrained local maximum. The circumstances under which the converse is true aregiven by the following theorem.

Theorem 49. Let X be a convex set inRn. Let f,g j( j = 1, ...,m) be concave functions onX. Supposex is a point of constrained local maximum. Then,x is a point of constrainedglobal maximum.

Proof. Sincex is a point of constrained local maximum, there isδ > 0, such that for allx ∈B(x,δ)∩C, we havef (x) ≤ f (x).

Now, if x is not a point of constrained global maximum, then there is some ¯x ∈C, suchthat f (x) > f (x). One can choose 0< θ < 1 with θ sufficiently close to zero, such thatx ≡ [θ x+ (1−θ)x] ∈ B(x,δ). SinceX is convex andg j( j = 1, ...,m) are concave,C is aconvex set, and ˜x≡ [θ x+(1−θ)x] ∈C. Thusx≡ [θ x+(1−θ)x] ∈ B(x,δ)∩C. Also, sincef is concave,f (x) = f (θ x+(1−θ)x) ≥ θ f (x)+(1−θ) f (x) > θ f (x)+(1−θ) f (x) = f (x).But this contradicts the fact that ˆx is a point of constrained local maximum.


10.6 Appendix: On the Kuhn-Tucker Conditions

Let X be an open set inRn, containingRn+, and f ,G j( j = 1, ...,r) be continuously differ-

entiable onX. We consider the optimization problem:

Max f(x)sub ject to Gj(x) ≥ 0 f or j = 1, ...,r

and x∈Rn+

⎫⎪⎪⎪⎬⎪⎪⎪⎭(P)We rewrite problem(P) in its equivalent form, given by:

Max f(x)sub ject to gj(x) ≥ 0 f or j = 1, ...,mand x∈X

⎫⎪⎪⎪⎬⎪⎪⎪⎭(Q)wherem= r +n, and:

g j(x) =G j(x) f or j = 1, . . . ,rg j(x) = x j−r f or j = r +1, . . . ,r +n

(1)

We can now write down the Kuhn-Tucker conditions for problem(Q). A pair (x, λ) inX×Rm

+ satisfies theKuhn-Tucker conditionsfor problem(Q) if:

(i) Di f (x)+ m∑j=1

λ jDig j(x) = 0 i = 1, ...,n

(ii) g(x) ≥ 0 and λg(x) = 0

⎫⎪⎪⎪⎬⎪⎪⎪⎭(KT I) (2)

Denotingλr+i by µi for i = 1, . . . ,n, andλ j = ν j for j = 1, . . . ,r, we see that (2) can bewritten as: (i) Di f (x)+ r∑

j=1ν jDiG j(x)+ µi = 0 i = 1, ...,n

(ii) G(x) ≥ 0, x≥ 0and νG(x) = 0, µx= 0(3)

Sinceµi ≥ 0 andµi xi = 0 for i = 1, . . . ,n by (3), we obtain(x, ν) ∈Rn+×Rr+:

(i)(a) Di f (x)+ r∑j=1

ν jDiG j(x) ≤ 0 i = 1, . . . ,n

(i)(b) [Di f (x)+ r∑j=1

ν jDiG j(x)]xi = 0 i = 1, . . . ,n

(ii)(a)G j(x) ≥ 0 f or j = 1, . . .r

(ii)(b) r∑j=1

ν jG j(x) = 0

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭(KT II) (4)


Conversely, given problem(P), suppose we can obtain(x, ν) ∈Rn+×Rr+ satisfying(KT

II ). Then we can define:

µi = −[Di f (x)+ r∑j=1

ν jDiG j(x)] f or i = 1, . . . ,n (5)

and:λ j = ν j f or j = 1, . . . ,r, λr+i = µi f or i = 1, . . . ,n (6)

Then denoting(r +n) by m, we obtainµ∈Rn+ by (KT II(i)(a)) and (5) andλ ∈Rm

+ by (6).Further, defining:

g j(x) =G j(x) f or j = 1, . . . ,rg j(x) = x j−r f or j = r +1, . . . ,r +n

(7)

we obtain:g j(x) ≥ 0 f or j = 1, . . .m (8)

sincex ∈Rn+ and(KT II(ii)(a)) holds.

Finally, using(KT II(i)(b)) and(KT II(ii)(b)), one obtains:

νG(x) = 0, µx= 0 (9)

Combining (5)-(9), we obtain(x, λ) in X×Rm+ such that:

(i) Di f (x)+ m∑j=1

λ jDig j(x) = 0 i = 1, ...,n

(ii) g(x) ≥ 0 and λg(x) = 0

⎫⎪⎪⎪⎬⎪⎪⎪⎭(KT I) (10)

Thus (KT I) and (KT II) are equivalent ways of writing the Kuhn-Tucker conditions,given problem(P).



Problem 41(Using Kuhn-Tucker Sufficiency Theory by Contracting the Domain).

Let p be an arbitrary positive real number. Consider the followingconstrained opti-mization problem:

Maximize x0.51 +x0.52

sub ject to px1+x2 ≤ px3+x4(x3)2+(x4)2 ≤ 1(x1,x2,x3,x4) ∈R4+

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭(R)

(a) To solve problem(R), first solve problem(S) given below:

Maximize x0.51 +x0.52

sub ject to px1+x2 ≤ px3+x4(x3)2+(x4)2 ≤ 1(x1,x2,x3,x4) ∈R4++

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭(S)

Define X = R4++ , f (x) = x0.51 + x0.5

2 , g1(x) = px3+ x4− px1− x2 , g2(x) = 1− [(x3)2+(x4)2] , wherex = (x1,x2,x3,x4) ∈ X. Write down and solve the Kuhn-Tucker conditionsfor problem(S), and denote the solution of the Kuhn-Tucker conditions by(x, λ) ∈X×R2+.

(b) Show that ¯x solves problem(S), and(x, λ) satisfies:

f (x)+ λg(x) ≤ f (x)+ λg(x) for all x ∈X

(c) Use (b) and the continuity off , g1 andg2 onR4+ to establish that ¯x solves(R).Solution.

(a) Define the setX =R4++ and define the functionsf , g1, andg2, each fromX to R, by

f (x) = x121 +x

122 for all x ∈X

g1(x) = px3+x4− px1−x2 for all x ∈X

g2(x) = 1−x23−x2

4 for all x ∈X

A pair (x, λ) ∈ X ×R2+ satisfies the Kuhn-Tucker conditions for problem(S) if itsatisfies the following:

12

x− 1

21 + λ1(−p) = 0 (1.1) px3+ x4− px1− x2 ≥ 0 (1.5)

12

x− 1

22 + λ1(−1) = 0 (1.2) 1− x2

3− x24 ≥ 0 (1.6)

λ1(p)+ λ2(−2x3) = 0 (1.3) λ1(px3+ x4− px1− x2) = 0 (1.7)λ1+ λ2(−2x4) = 0 (1.4) λ2(1− x2

3− x24) = 0 (1.8)


Since the objective function is increasing inx1 andx2, we expect that the first con-straint in problem(S)will hold with equality at any solution. In addition, the greateris the right hand side of the first constraint, the greater is the value of the objectivefunction that we can achieve. So we expect the second constraint in problem(S) tohold with equality, also.

Seeking contradiction, suppose thatλ1 = 0. Then (1.1) and (1.2) cannot hold, so wehave a contradiction. Thereforeλ1 > 0. By (1.7), then, we have

px3+ x4− px1− x2 = 0 (1.9)

Again seeking contradiction, suppose thatλ2 = 0. But then (1.4) implies thatλ1 = 0,which contradicts what we have just shown. Thereforeλ2 > 0, so by (1.8) we have

1− x23− x2

4 = 0 (1.10)

From (1.3) and (1.4), we have thatp= x3x4

. Using this in (1.10), we have that(px4)2+x2

4 = 1, which gives

x4 =1√

1+ p2, x3 = px4 =

p√1+ p2

Now, using (1.1) and (1.2) to eliminateλ1, we havep=√

x2x1

, or x2 = p2x1. Then we

can use (1.9) andp= x3x4

to write p2x4+ x4− px1− p2x1 = 0. This implies that

x1 =1+ p2

p(1+ p) x4 =

√1+ p2

p(1+ p) , x2 = p2x1 =p√

1+ p2

1+ p

From (1.2) and then (1.4), we have

λ1 =12

x− 1

22 =

(1+ p) 12

2p12(1+ p2) 1

4

, λ2 =λ1

2x4=(1+ p) 1

2(1+ p2) 14

4p12

The pair(x, λ) ∈X×R2+ given above is the unique solution to the Kuhn-Tucker con-ditions for problem(S).


(b) We need to verify that the conditions of the Kuhn-Tucker Sufficiency Theorem aresatisfied before we can conclude that ¯x solves problem(S).The setX is open and convex inR4. The functionsf , g1, andg2 are continuouslydifferentiable onX.

The function f can be expressed as the sum of functions that are concave onX, sof is concave onX. The functiong1 is linear onX, so it is concave onX. Now, thefunctionsh1(x) = x2

3 andh3(x) = x24 are convex onX, so−h1 and−h2 are concave on

X. Since we can writeg2(x) = 1−h1(x)−h2(x) for all x ∈ X, we can expressg2 asthe sum of functions that are concave onX. This means thatg2 is concave onX.

So all the conditions of the Kuhn-Tucker Sufficiency Theoremare met. Since(x, λ) ∈X×R2+ satisfies the Kuhn-Tucker conditions for problem(S), then,x solves problem(S).Theorem 46, Chapter 10, which is the Kuhn-Tucker Sufficiency Theorem, includesthe result that(x, λ) is a saddle point. By the definition of a saddle point, then, wehave thatf (x)+ λg(x) ≤ f (x)+ λg(x) for all x ∈X.

(c) Let x be an arbitrary point inR4+, and define the sequence{xn}∞n=1 by

xn = (x1+ 1n,x2+ 1

n,x3+ 1

n,x4+ 1

n) for all n= 1,2, . . .

Thenxn ∈ R4++ for all n = 1,2, . . . , and the sequence{xn}∞n=1 converges tox. Sincef , g1, andg2 are continuous onR4+, the sequence{ f (xn)}∞n=1 converges tof (x), thesequence{g1(xn)}∞n=1 converges tog1(x), and the sequence{g2(xn)}∞n=1 convergesto g2(x). That is, the sequence{ f (xn)+ λg(xn)}∞n=1 converges tof (x)+ λg(x).Using the result in part (b) and the fact thatxn ∈ X = R4++ for all n = 1,2, . . . , wehave thatf (xn)+ λg(xn) ≤ f (x)+ λg(x) for all n= 1,2, . . . . Since weak inequalitiesare preserved in the limit and the sequence{ f (xn)+ λg(xn)}∞n=1 converges tof (x)+λg(x), we have thatf (x)+ λg(x) ≤ f (x)+ λg(x) for our arbitraryx ∈ R4+, and thusfor all x ∈R4+.

Now, in part (b) we said that(x, λ) is a saddle point for problem(S), which includesthe inequality f (x)+ λg(x) ≤ f (x)+λg(x) for all λ ∈ R2+. So, from this and theprevious paragraph, we have that

f (x)+ λg(x) ≤ f (x)+ λg(x) ≤ f (x)+λg(x) for all (x,λ) ∈R4+×R2

+

That is,(x, λ) is a saddle point for problem(R). By Theorem 44, then, ¯x solvesproblem(R).


Problem 42(Using Kuhn-Tucker Sufficiency Theory by Expanding the Domain).

Let a,b be arbitrary positive numbers, satisfyinga > b > 1. Consider the following con-strained maximization problem:

Maximize aln(1+x1)+bln(1+x2)+x3

sub ject to x1+x2+x3 ≤ 1and (x1,x2,x3) ∈R3

+

⎫⎪⎪⎪⎬⎪⎪⎪⎭(R)(a) DefineX = {(x1,x2,x3) ∈R3 ∶ xi > −1 for all i ∈ {1,2,3}}. Now, definef (x1,x2,x3) =

aln(1+x1)+bln(1+x2)+x3 , g1(x1,x2,x3)=1−(x1+x2+x3), g2(x1,x2,x3)=x1, g3(x1,x2,x3)=x2, g4(x1,x2,x3) = x3 for all (x1,x2,x3) ∈X. Write down the appropriate Kuhn-Tucker con-ditions for problem(R).

(b) Solve the Kuhn-Tucker conditions in each of the following three cases: (i)a≥ 2b ;(ii) a< 2b and(a+b) ≥ 3 ; (iii) a< 2b and(a+b) < 3.

(c) Use your solutions to the Kuhn-Tucker conditions to obtain solutions to(R) in eachof the three cases specified in (b) above.

Solution.

(a) Define the set

X = {(x1,x2,x3) ∈R3 ∣ xi > −1 for all i = 1,2,3}Now, define the following functions, each fromX to R:

f (x1,x2,x3) = aln(1+x1)+bln(1+x2)+x3 for all (x1,x2,x3) ∈X

g1(x1,x2,x3) = 1−(x1+x2+x3) for all (x1,x2,x3) ∈X

g2(x1,x2,x3) = x1 for all (x1,x2,x3) ∈X



A pair (x, λ) ∈ X ×R4+ satisfies the Kuhn-Tucker conditions for problem(R) if it


satisfies the following:

a1+ x1

− λ1+ λ2 = 0 (2.1) x3 ≥ 0 (2.7)b

1+ x2− λ1+ λ3 = 0 (2.2) λ1[1−(x1+ x2+ x3)] = 0 (2.8)

1− λ1+ λ4 = 0 (2.3) λ2x1 = 0 (2.9)1−(x1+ x2+ x3) ≥ 0 (2.4) λ3x2 = 0 (2.10)

x1 ≥ 0 (2.5) λ4x3 = 0 (2.11)x2 ≥ 0 (2.6)

(b) We are looking for(x, λ) ∈ X ×R4+ satisfying (2.1)–(2.11). The motivation for thefollowing strategy was discussed in the final section meeting of the course.

First, we want to show that the budget constraint, (2.4), holds with equality. Seekingcontradiction, suppose thatλ1 = 0. Then (2.3) givesλ4 = −1≱ 0, which is a contra-diction. Soλ1 > 0. Then by (2.8), we have

1−(x1+ x2+ x3) = 0 (2.12)

Next, we want to show that ˆx1 > 0. Seeking contradiction, suppose ˆx1 = 0. Then by(2.1), λ1 = a+ λ2 ≥ a > 1. Then (2.3) implies thatλ4 = λ1−1 > 0, so (2.11) impliesthat x3 = 0. Then from (2.12) we have ˆx2 = 1, so (2.10) givesλ3 = 0. By (2.2), then,λ1 =

b2, so by (2.1) we haveλ2 =

b2 −a. But we requireλ2 ≥ 0, which can only hold

if b≥ 2a. This contradicts the given information thata> b> 1, so we conclude thatx1 > 0.

At this point we need to split the problem into cases.

Case 1: ˆx1 > 0, x2 = 0. By (2.9), we have thatλ2 = 0. Since ˆx2 = 0, it follows from(2.2) thatb− λ1+ λ3 = 0. Then (2.3) givesλ4 = λ1−1= b+ λ3−1> 0 sinceb> 1, so(2.11) implies ˆx3 = 0. From (2.12), then, we have ˆx1 = 1. Plugging this into (2.1) andusing the fact thatλ2 = 0, we haveλ1 =

a2. Then (2.2) yieldsλ3 =

a2 −b. Since we

requireλ3 ≥ 0, we must havea≥ 2b. From (2.3) we haveλ4 =a2−1. Whena≥ 2b we

have thata> 2, soλ4 > 0.

Case 2: ˆx1 > 0, x2 > 0, x3 = 0. By (2.9) and (2.10), we have thatλ2 = 0 andλ3 = 0.By (2.12), we have ˆx1+ x2 = 1. Using this after solving (2.1) and (2.2) forλ1 andequating the two expressions, we havea2−x2

= b1+x2

. This gives ˆx2 =2b−aa+b . Since we

must have ˆx2 > 0, this case requiresa< 2b. Thenx1 = 1− x2 =2a−ba+b > 0. Next, we can


use (2.1) to solve forλ1 =a

1+x1= a+b

3 . Then from (2.3) we haveλ4 = λ1−1= a+b3 −1.

Since we must haveλ4 ≥ 0, a further requirement for this case isa+b≥ 3.

Case 3: ˆx1 > 0, x2 > 0, x3 > 0. By (2.9), (2.10), and (2.11), we have thatλ2 = 0,λ3 = 0, andλ4 = 0. Then (2.3) yieldsλ1 = 1. Using this in (2.1) and (2.2), we havex1 = a−1> 0 andx2 = b−1> 0. Then (2.12) implies ˆx3 = 1− x1− x2 = 3−(a+b), sox3 > 0 requiresa+b< 3. Now, if a+b< 3 anda> b> 1, it must also be thata< 2b.

To summarize:

(i) If a≥ 2b, the unique solution to the Kuhn-Tucker conditions is

(x, λ) = ((1,0,0) ,(a2,0,

a2−b,

a2−1))

(ii) If a< 2b anda+b≥ 3, the unique solution to the Kuhn-Tucker conditions is

(x, λ) = ((2a−ba+b

,

2b−aa+b

,0) ,(a+b3

,0,0,a+b

3−1))

(iii) If a< 2b anda+b< 3, the unique solution to the Kuhn-Tucker conditions is

(x, λ) = ((a−1,b−1,3−(a+b)),(1,0,0,0))(c) We need to verify that the conditions of the Kuhn-Tucker Sufficiency Theorem are

satisfied before we can conclude that the various ˆxwe found in part (b) solve problem(R) under the associated parameter restrictions.

The setX, defined in part (a), is open and convex inR3. The functionsf , g1, g2, g3,andg4 are continuously differentiable onX.

For allx ∈X, the Hessian off atx is

H f (x) =⎡⎢⎢⎢⎢⎢⎢⎣− a(1+x1)2 0 0

0 − b(1+x2)2 0

0 0 0

⎤⎥⎥⎥⎥⎥⎥⎦The first order principal minors ofH f (x) are− a

(1+x1)2 , − b(1+x2)2 , and 0, which are

each less than or equal to zero for allx ∈ X. The second order principal minors ofH f (x) are a

(1+x1)2b

(1+x2)2 and 0, which are both greater than or equal to zero for all


x ∈ X. The determinant ofH f (x) is zero for allx ∈ X. SoH f (x) is negative semi-definite for allx ∈X. Therefore, sinceX is an open set, we have thatf is concave onX.

Since the functionsg1, g2, g3, andg4 are linear onX, they are concave onX.

By the Kuhn-Tucker Sufficiency Theorem, if a pair(x, λ) ∈X×R4+ satisfies the Kuhn-Tucker conditions given by (2.1)–(2.11), then ˆx solves problem(R). So from part(b) we have the following solutions to problem(R), depending on the values of theparametersa andb:

(i) If a≥ 2b, thenx= (1,0,0) solves(R).(ii) If a< 2b anda+b≥ 3, then ˆx= (2a−b

a+b ,2b−aa+b ,0) solves(R).

(iii) If a< 2b anda+b< 3, then ˆx= (a−1,b−1,3−(a+b)) solves(R).Problem 43 (Using Kuhn-Tucker Sufficiency Theory for a Non-Differentiable ObjectiveFunction).

Let a1,a2, p1, p2,w be arbitrary positive numbers. Consider the constrained optimizationproblem:

Maximize min{a1x1,a2x2}sub ject to p1x1+ p2x2 ≤w(x1,x2) ≥ 0

⎫⎪⎪⎪⎬⎪⎪⎪⎭(Q)Here the objective function is not differentiable (the indifference curves are L-shaped).Instead of solving(Q), we consider the following constrained optimization problem:

Maximize a1x1

sub ject to a2x2 ≥ a1x1

p1x1+ p2x2 ≤w(x1,x2) ≥ 0

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭(Q′)

(a) Solve problem(Q′) by using the Kuhn-Tucker sufficiency theorem.(b) Show that any solution of(Q′) is also a solution of(Q).

Solution.

(a) To use the Kuhn-Tucker Sufficiency Theorem, we need to specify an open setX.The setR2++ would work, but we would have to go through the trouble of ruling outcorner solutions. On the other hand, if we think about solving the utility maximiza-tion problem on all ofR2, it is clear that the solution will be somewhere inR2++. So


it suffices to useX = R2, and we don’t even have to include the constraintsx1 ≥ 0andx2 ≥ 0 as long as we find a solution ˆx ∈R2+ for all positive parameter values, asrequired by the problem.

Define the following functions, each fromX =R2 toR:

f (x1,x2) = a1x1 for all (x1,x2) ∈X

g1(x1,x2) = a2x2−a1x1 for all (x1,x2) ∈X

g2(x1,x2) =w− p1x1− p2x2 for all (x1,x2) ∈X

Now, X is open and convex inR2. The functionsf , g1, andg2 are each linear onX, so they are each continuously differentiable and concave on X. Therefore all theconditions of the Kuhn-Tucker Sufficiency Theorem are met. Apair (x, λ) ∈X×R2+satisfies the Kuhn-Tucker conditions for problem(Q′) if it satisfies the following:

a1− λ1a1− λ2p1 = 0 (3.1)

λ1a2− λ2p2 = 0 (3.2)

a2x2−a1x1 ≥ 0 (3.3)

w− p1x1− p2x2 ≥ 0 (3.4)

λ1(a2x2−a1x1) = 0 (3.5)

λ2(w− p1x1− p2x2) = 0 (3.6)

First, note that (3.1) and (3.2) form two equations in two unknowns. Without divid-ing by λ1 or λ2, we can solve them for

λ1 =a1p2

a2p1+a1p2, λ2 =

a1a2

a2p1+a1p2

Sinceλ1 > 0 andλ2 > 0, from (3.5) and (3.6) we have thata2x2−a1x1 = 0 andw−p1x1− p2x2 = 0. This is again two equations in two unknowns, and without dividingby x1 or x2 we can solve them for

x1 =a2w

a2p1+a1p2> 0, x2 =

a1wa2p1+a1p2

> 0

By the Kuhn-Tucker Sufficiency Theorem,(x1, x2) solves problem(Q′).


(b) Consider the problem

maximize a2x2

subject to a1x1 ≥ a2x2

p1x1+ p2x2 ≤w(x1,x2) ≥ 0

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭(Q′′)

Following steps symmetric to those in part (a), we have that the solution ˆx also solvesproblem(Q′′).Let x by a solution to problem(Q′). Thenx also solves problem(Q′′). Note thata1x1 = a2x2. Seeking contradiction, suppose that ¯x is not a solution to(Q). Thenthere is somex′ ∈R2+, x′ ≠ x such thatp1x′1+ p2x′2 ≤w and min{a1x′1,a2x′2} > a1x1 =

a2x2. We can consider two cases.

Case 1: min{a1x′1,a2x′2} = a1x′1. Then we havea1x′1 ≤ a2x′2, so thatx′ is in theconstraint set of problem(Q′), anda1x′1 > a1x1. This contradicts ¯x solving problem(Q′).Case 2: min{a1x′1,a2x′2} = a2x′2. Then we havea2x′2 ≤ a1x′1, so thatx′ is in theconstraint set of problem(Q′′), anda2x′2 > a2x2. This contradicts ¯x solving problem(Q′′).In either of these two collectively exhaustive cases, we arrive at a contradiction. Soit must be that if ¯x solves(Q′), thenx also solves(Q).

Problem 44(Applying the Arrow-Hurwicz-Uzawa Necessity Theorem).

Let a,b,c andp,q,r be arbitrary positive numbers, satisfying: 1> (p/r) ≥ (a+b)c. Con-sider the following two problems of constrained optimization:

Maximize xac1 xbc

2 − px1−qx2

sub ject to (x1,x2) ∈R2+}(Q)

Maximize xac1 xbc

2 − px1−qx2

sub ject to rx1+qx2−xac1 xbc

2 ≥ 0(x1,x2) ∈R2+

⎫⎪⎪⎪⎬⎪⎪⎪⎭(R)(a) Using an appropriately modified version of Weierstrass theorem, establish that there

is a solution ¯x to problem(Q), and a solution ˆx to problem(R).(b) Show that ¯xi > 0 for i = 1,2 , and ˆxi > 0 for i = 1,2.(c) Use the Arrow-Hurwicz-Uzawa necessity theorem to compare the solutions ¯x and

x , and show that: (x1/ x2) > (x1/ x2)


explaining your procedure clearly.[Hint: you might want touse the fact thatxac1 xbc

2 ishomogeneous of degree(a+b)c in (x1,x2)].Solution.

(a) Define f ∶R2+→R andg∶R2+→R by

f (x1,x2) = xac1 xbc

2 − px1−qx2 for all (x1,x2) ∈R2+

g(x1,x2) = rx1+qx2−xac1 xbc

2 for all (x1,x2) ∈R2+

Note thatf andg are continuous onR2+. Define the constraint sets

CQ =R2+, CR= {(x1,x2) ∈R2

+ ∣ g(x1,x2) ≥ 0}Now,CQ is closed inR2 sinceR2+ is closed inR2, andCQ is nonempty since(0,0) ∈CQ. Also,CR is closed inR2 sinceg is continuous onR2+, andCR is nonempty since(0,0) ∈CR. ButCQ andCR are both unbounded sets inR2.

Because(a+b)c < 1, for large enoughx1 or x2, we will have f (x1,x2) < 0. Sowe can use(0,0), with f (0,0) = 0, as our comparison point in order to apply theExtension of Weierstrass Theorem. Here is a formal argumentof this. First, definek=max{x1,x2} andπ =min{p,q}. Then we have

f (x1,x2) = xac1 xbc

2 − px1−qx2

≤ [max{x1,x2}](a+b)c− [min{p,q}](x1+x2)= k(a+b)c−π(x1+x2)≤ k(a+b)c−π[max{x1,x2}]= k(a+b)c−πk

Define the functione∶R+→R by

e(k) = k(a+b)c−πk for all k≥ 0

Our goal is to find somek> 0 such thate(k) < 0 wheneverk> k. That will imply thatf (x1,x2) < 0 whenever max{x1,x2} > k.

For k > 0, there is a unique solution to the equatione(k) = 0, which isk = π1

(a+b)c−1 .Now, e′(k) = (a+b)ck(a+b)c−1−π for all k > 0. Since(a+b)c−1< 0, we have thate′(k) = π((a+b)c−1) < 0. Sincek is the unique solution toe(k) = 0 for k > 0, e is


continuous on its domain, ande′(k) <0, we have thate(k) <0 for all k> k. Thereforef (x1,x2) < 0 whenever max{x1,x2} > k.

Define the sets

CQ = {(x1,x2) ∈R2+ ∣max{x1,x2} ≤ k}

CR= {(x1,x2) ∈R2+ ∣max{x1,x2} ≤ k andg(x1,x2) ≥ 0}

Since (0,0) ∈ CQ and (0,0) ∈ CR, these sets are nonempty. Sinceg(x1,x2) andmax{x1,x2} are continuous functions onR2+ and the sets are defined by weak in-equalities, the sets are closed. And the restriction max{x1,x2} ≤ k means that thesets are bounded.

Now, for any(x1,x2) ∈CQ∖CQ, or for any(x1,x2) ∈CR∖CR, we have that max{x1,x2}>k, so it must be thatf (x1,x2) < 0= f (0,0). So by the Extension of Weierstrass The-orem, there is some ¯x ∈ CQ that solves problem(Q) and some ˆx ∈ CR that solvesproblem(R).

(b) Seeking contradiction, suppose that ¯x1 = 0 or x2 = 0. Then f (x1, x2) ≤ 0. But for

0< ε < (p+q) 1(a+b)c−1 , we have that

f (ε,ε) = εacεbc− pε−qε

= ε(ε(a+b)c−1− p−q)> 0

Since(ε,ε) ∈CQ and f (ε,ε)> f (x1, x2), we have a contradiction of ¯xsolving problem(Q). So it must be that ¯x1 > 0 andx2 > 0.

The idea for showing that ˆx1 > 0 andx2 > 0 is conceptually similar, but the proof is abit trickier because we now have the constraint to worry about. First, define

δ = (r +q) 1(a+b)c−1 > 0

This ensures that the constraint is satisfied:g(δ,δ) = 0. Seeking contradiction, sup-pose that ˆx1 = 0 or x2 = 0. Then f (x1, x2) ≤ 0. Now, sincer > p and(a+b)c< 1, wehave that

δ = (r +q) 1(a+b)c−1 < (p+q) 1

(a+b)c−1

Thereforef (δ,δ) > 0. Then(δ,δ) ∈CR and f (δ,δ) > f (x1, x2), which is a contradic-tion of x solving problem(R). So it must be that ˆx1 > 0 andx2 > 0.


(c) DefineX =R2++. ThenX is open inR2 and the functionsf andg are continuouslydifferentiable onX. By parts (a) and (b), there is some ¯x ∈ X that is a point oflocal maximum in problem(Q) and some ˆx ∈X that is a point of local maximum inproblem(R).We need to verify that the Arrow-Hurwicz-Uzawa Constraint Qualification is satis-fied. We have thatX is convex. For allx ∈X, the Hessian ofg atx is

Hg(x) = [ ac(1−ac)xac−21 xbc

2 abc2xac−11 xbc−1

2abc2xac−1

1 xbc−12 bc(1−bc)xac

1 xbc−22

]Since(a+b)c < 1 with each parameter positive, it must be thatac< 1 andbc< 1.Thereforeac(1−ac)xac−2

1 xbc2 > 0 andbc(1−bc)xac

1 xbc−22 > 0. Now, the determinant

of the Hessian ofg at anyx∈X is abc2(1−ac−bc)x2ac−21 x2bc−2

2 > 0 since(a+b)c< 1.This means that the Hessian ofg is positive semi-definite for allx ∈ X, which isequivalent to saying thatg is a convex function onX. Therefore the Arrow-Hurwicz-Uzawa Constraint Qualification is satisfied.

By the Arrow-Hurwicz-Uzawa Necessity Theorem, then, there is someλ ∈R+ suchthat x satisfies the Kuhn-Tucker conditions for problem(Q) and(x, λ) satisfies theKuhn-Tucker conditions for problem(R).Note that because of the result in part (b), we could apply theArrow-Hurwicz-Uzawa Necessity Theorem to problem(R) only and use unconstrained optimizationtheory on problem(Q). This would require showing thatf is concave onR2++, sothat the first order conditions are sufficient for a solution.

Also note that regarding problem(R), the result in part (b) allows us to leave theconstraints ˆx1 > 0 andx2 > 0 out of the Kuhn-Tucker conditions, so that we do nothave to deal with two additional multipliers.

Now, the Kuhn-Tucker conditions for problem(Q) are

acxac−11 xbc

2 − p= 0 (4.1)

bcxac1 xbc−1

2 −q= 0 (4.2)

From (4.1) and (4.2), we can obtain

x1

x2=

qp

ab

(4.3)


The Kuhn-Tucker conditions for problem(R) are

acxac−11 xbc

2 − p+ λ(r −acxac−11 xbc

2 ) = 0 (4.4)

bcxac1 xbc−1

2 −q+ λ(q−bcxac1 xbc−1

2 ) = 0 (4.5)

rx1+qx2− xac1 xbc

2 ≥ 0 (4.6)

λ(rx1+qx2− xac1 xbc

2 ) = 0 (4.7)

If λ = 0 then the solution to problem(R) will be the same as the solution to problem(Q). But we wantx1x2> x1

x2, so we need to show thatλ ≠ 0.

We can start by rewriting (4.5) as

(λ−1)(q−bcxac1 xbc−1

2 ) = 0 (4.8)

Now, if λ = 1, then from (4.4) we haver = p, which is a contradiction of the giveninformation thatr > p. So it must be thatλ ≠ 1. Then by (4.8), we have

qx2 = bcxac1 xbc

2 (4.9)

Seeking contradiction, supposeλ = 0. Then by (4.4), we havepx1 = acxac1 xbc

2 . Using(4.9) in (4.6), we haverx1 ≥ (1−bc)xac

1 xbc2 . Now, divide the left side of this inequal-

ity by px1 and the right side byacxac1 xbc

2 , recalling that the divisors are equal. Theresult is r

p ≥1−bc

ac . But we are also given thatrp ≤

1(a+b)c, so we can form the inequality

1(a+b)c ≥

1−bcac , which is equivalent toa≥ (1−bc)(a+b) = a+b−abc−b2c. Rearrang-

ing, we have thatb(1−ac−bc) ≤ 0, which, sinceb> 0, implies that(a+b)c≥ 1. Butthis is a contradiction of the given information that(a+b)c< 1, so we can concludethatλ > 0.

Using (4.7) and (4.9), then, we have

rx1+(bc−1)xac1 xbc

2 = 0 (4.10)

Since(a+b)c < 1, we havebc−1 < −ac, so (4.10) implies thatrx1−acxac1 xbc

2 > 0.Multiplying (4.4) by x1, then, we have

acxac1 xbc

2 < px1 (4.11)

We can divide the left side of (4.11) bybcxac1 xbc

2 and the right side byqx2. Since thedivisors are equal by (4.9), we have

x1

x2>

qp

ab

(4.12)


Combining (4.3) and (4.12), we have the result:

x1

x2>

x1

x2


References:This material is based onNon-Linear Programmingby O.L. Mangasarian(Chapters

5, 7); Mathematical Economicsby A. Takayama(Chapter 1); andConvex Structures andEconomic Theoryby H. Nikaido(Chapter 1).

The proof of Theorem 45 (the Kuhn-Tucker theorem), relying on Minkowski’s sepa-ration theorem and Slater’s condition is due to H. Uzawa: “The Kuhn-Tucker Theorem inConcave Programming” inStudies in Linear and Non-Linear Programming(1958), editedby K.J. Arrow, L. Hurwicz and H. Uzawa. For more on Theorem 48 (the Arrow-Hurwicz-Uzawa theorem), see K.J. Arrow, L. Hurwicz and H. Uzawa, “Constraint Qualifications inMaximization Problems”,Naval Research Logistic Quarterly, 8 (1961), 175-191.

Chapter 11

Quasi-Concave Programming

11.1 Properties of Concave and Quasi-concave functions

For this section, it is convenient to denoteRn+ by Y andRn

++ by Z.

11.1.1 Concave Functions

Lemma 1. If h ∶Y→R is continuous on Y and concave on Z, then it is concave on Y.

Proof. Let x, x ∈Y and 0< θ < 1 be given. Denote the vector(1, . . . ,1) ∈Y by u and define:

xs= x+(u/s), xs= x+(u/s) f or s= 1,2,3, ... (11.1)

Thenxs ∈ Z andxs ∈ Z for eachs. Sinceh is concave onZ, we have for eachs:

h(θxs+(1−θ)xs) ≥ θh(xs)+(1−θ)h(xs) (11.2)

Now, lets→∞. Then,xs→ x, xs

→ x and(θxs+(1−θ)xs)→ (θx+(1−θ)x). Using (11.2)and the continuity ofh onY, we get:

h(θx+(1−θ)x) ≥ θh(x)+(1−θ)h(x) (11.3)

This establishes thath is concave onY.

Lemma 2. Let X be an open set containing Y. Let h be continuously differentiable on Xand concave on Y. If x, x ∈Y, then:

h(x)−h(x) ≤ (x− x)∇h(x) (11.4)

181

CHAPTER 11. QUASI-CONCAVE PROGRAMMING 182

Proof. Denote the vector(1, . . . ,1) ∈Y by u and define:

xs= x+(u/s), xs= x+(u/s) f or s= 1,2,3, ... (11.5)

Thenxs ∈Z andxs ∈Z for eachs. Sinceh is concave and continuously differentiable on theopen convex setZ, we have for eachs:

h(xs)−h(xs) ≤ (xs− xs)∇h(xs) (11.6)

Now, let s→∞. Then,xs→ x, xs

→ x. Using (11.6) and the continuous differentiability ofh onX, we get (11.4).

11.1.2 Quasi-concave functions

Lemma 3. If h ∶Y→R is continuous on Y and quasi-concave on Z, then it is quasi-concaveon Y.

Proof. Let x, x ∈Y be given withh(x) ≥ h(x), and let 0< θ < 1 be given. We have to showthat:

h(θx+(1−θ)x) ≥ h(x) (11.7)

Suppose, contrary to (11.7), we have:

h(θx+(1−θ)x) < h(x) (11.8)

Then, there isε > 0 such that:

h(θx+(1−θ)x) < h(x)−ε (11.9)

Denote the vector(1, . . . ,1) ∈Y by u and define:

xs= x+(u/s), xs= x+(u/s) f or s= 1,2,3, ... (11.10)

Then, sincexs→ x andxs

→ x ass→∞, andh is continuous onY, we can findSsuch thatfor all s≥S,

h(xs) ≥ h(x)−ε, h(xs) ≥ h(x)−ε (11.11)

Sinceh is quasi-concave onZ, andxs ∈ Z, xs ∈ Z for s≥S, we get:

h(θxs+(1−θ)xs) ≥ min{h(xs),h(xs)}≥ min{h(x)−ε,h(x)−ε}= h(x)−ε (11.12)

Since(θxs+(1−θ)xs)→ (θx+(1−θ)x) ass→∞, andh is continuous onY, (12) yields:

h(θx+(1−θ)x) ≥ h(x)−ε

which contradicts (11.9) and establishes (11.7).


Lemma 4. Let X be an open set containing Y. Let h be continuously differentiable on Xand quasi-concave on Y. If x, x ∈Y, and h(x) ≥ h(x), then:

(x− x)∇h(x) ≥ 0 (11.13)

Proof. For 0< θ < 1, definex(θ) = (θx+(1−θ)x); thenx(θ) ∈Y, sinceY is a convex set.By quasi-concavity ofh onY,

h(x(θ)) = h(θx+(1−θ)x) ≥ h(x) (11.14)

Sinceh is continuously differentiable onX, we can apply the Mean-Value Theorem toobtainz(θ) ∈Y, such that:

h(x(θ))−h(x) = [x(θ)− x]∇h(z(θ)) = θ[x− x]∇h(z(θ)) (11.15)

wherez(θ) is a convex combination ofx(θ) andx. Using (11.14) and (11.15), we obtainfor each 0< θ < 1: [x− x]∇h(z(θ)) ≥ 0 (11.16)

Lettingθ→ 0, we see thatx(θ)→ x and soz(θ)→ x. Using the continuous differentiabilityof h onX,and (11.16), we get (11.13).

11.2 Definitions

Let X be an open set inRn, containingRn+, and f ,G j( j = 1, ...,r) be continuously differ-

entiable onX. We are concerned with the optimization problem:

Max f(x)sub ject to Gj(x) ≥ 0 f or j = 1, ...,r

and x∈Rn+

⎫⎪⎪⎪⎬⎪⎪⎪⎭(P)A pair (x, ν) ∈Rn

+×Rr+ satisfies theKuhn-Tucker conditionsif:

(i)(a) Di f (x)+ r∑j=1

ν jDiG j(x) ≤ 0 i = 1, . . . ,n

(i)(b) [Di f (x)+ r∑j=1

ν jDiG j(x)]xi = 0 i = 1, . . . ,n

(ii)(a) G j(x) ≥ 0 f or j = 1, . . .r

(ii)(b) r∑j=1

ν jG j(x) = 0

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭(KT II) (11.17)


For problem(P), these Kuhn-Tucker conditions(KT II) are equivalent to the ones in-troduced in our discussion of concave programming in Chapter10 (see handout on Kuhn-Tucker conditions).

The constraint setis defined asC = {x ∈ Rn+ ∶G j(x) ≥ 0 for j = 1, ...,r}. We say that

Slater’s Conditionis satisfied if there isx∗ ∈C, such thatG j(x∗) > 0 for j = 1, ...,r.

11.3 The Sufficiency Theorem of Arrow-Enthoven

Lemma 5. Suppose f,G j( j =1, ...,r) are continuously differentiable functions on X. Sup-pose there is a pair(x, ν) ∈Rn

+×Rr+, such that(x, ν) satisfies the Kuhn-Tucker conditions.If each Gj is quasi-concave onRn

+, then

x ∈C implies(x− x)∇ f (x) ≤ 0 (11.18)

Proof. Let x ∈C. Then, we have

(x− x)∇ f (x) = (x− x)[∇ f (x)+ r∑j=1

ν j∇G j(x)]−(x− x) r∑

j=1ν j∇G j(x)

= x[∇ f (x)+ r∑j=1

ν j∇G j(x)]−(x− x) r∑

j=1ν j∇G j(x)

≤ −(x− x) r∑j=1

ν j∇G j(x)If ν j = 0 for some j, then ν j(x− x)∇G j(x) = 0 for that j. If ν j > 0 for some j, thenG j(x) = 0 for that j, soG j(x)−G j(x) =G j(x) ≥ 0. By quasi-concavity ofG j , we thenhaveν j(x− x)∇G j(x) ≥ 0 for that j, by Lemma 4. Thus, for eachj, ν j(x− x)∇G j(x) ≥ 0.Consequently, we have(x− x)∇ f (x) ≤ 0.

To state the sufficiency result of Arrow and Enthoven, let us call an indexk ∈ {1, ...,n}a relevant indexif there existsx∗ ∈C, such thatx∗k > 0. We defineI as the set of relevantindices.

Theorem 50. Suppose f, G j( j = 1, ...,r) are continuously differentiable on X, and quasi-concave onRn

+. Suppose there is a pair(x, ν) ∈ Rn+ ×Rr+, such that(x, ν) satisfies the


Kuhn-Tucker conditions. Suppose, further that at least oneof the following conditions issatisfied:

(a) Di f (x) < 0 for some i∈ [1, ...,n].(b) Di f (x) > 0 for some i∈ I(c) f is concave on X.

Thenx solves(P).Proof. If condition (a) holds, there is some indexk, such thatDk f (x) < 0. Let ek be thekth unit vector, and define ˜x= x+ek. Thenx ∈Rn

+, and

(x− x)∇ f (x) < 0 (11.19)

Let x be an arbitrary vector inC. We have to show thatf (x) ≤ f (x). To this end, definefor 0< θ < 1, x(θ) = (1−θ)x+θ x,y(θ) = (1−θ)x+θx. Then, usingθ > 0 and (11.19), wehave [y(θ)− x]∇ f (x) = θ(x− x)∇ f (x) < 0 (11.20)

Also, by Lemma 5,

[x(θ)−y(θ)]∇ f (x) = (1−θ)(x− x)∇ f (x) ≤ 0 (11.21)

Thus, adding (11.20) and (11.21), we get[x(θ)− x]∇ f (x) < 0. Sincef is quasi-concave,we havef (x(θ)) < f (x) by Lemma 4. Lettingθ→ 0, we getf (x) ≤ f (x).

Suppose condition (b) holds. If condition (a) still holds, we are already done. So,assume that (a) does not hold. That is∇ f (x) ≥ 0, andDk f (x) > 0 for some indexk ∈ I .Thus, there isx∗ ∈C such thatx∗k > 0. Using Lemma 5,

x∇ f (x) ≥ x∗∇ f (x) > 0 (11.22)

Let x ∈C. Then for 0< θ < 1, by using Lemma 5 and (11.22), we get(θ x)∇ f (x) ≤θ x∇ f (x) < x∇ f (x). Using the quasi-concavity off and Lemma 4, we havef (θ x) < f (x).Letting θ→ 1, we getf (x) ≤ f (x).

Suppose condition (c) holds. If ¯x∈C, we havef (x)− f (x) ≤ (x− x)∇ f (x) ≤ 0, by usingLemmas 2 and 5, and concavity off .

Corollary 8. Suppose f , Gj( j = 1, ...,r) are continuously differentiable on X, and quasi-concave onRn

+. Suppose there is a pair(x, ν) ∈ Rn+ ×Rr+, such that(x, ν) satisfies the

Kuhn-Tucker conditions. Suppose there is x∗ >> 0, such that Gj(x∗) ≥ 0 for j = 1, ...,r, and∇ f (x) ≠ 0. Thenx solves(P).


Proof. Since there isx∗ >> 0, such thatG j(x∗) ≥ 0 for j = 1, ...,r, all indicesi ∈ {1, ...,n}are relevant indices. Since∇ f (x) ≠ 0, there is some indexk for which Dk f (x) ≠ 0. IfDk f (x) < 0 then condition (a) of Theorem 50 is satisfied. IfDk f (x) > 0, then condition (b)of Theorem 50 is satisfied. Thus, in either case, the result follows from Theorem 50.

Corollary 9. Suppose f , Gj( j = 1, ...,r) are continuously differentiable on X, and quasi-concave onRn

+. Suppose there is a pair(x, ν) ∈ Rn+ ×Rr+ such that(x, ν) satisfies the

Kuhn-Tucker conditions. Suppose Slater’s condition is satisfied, and∇ f (x) ≠ 0. Thenxsolves(P).Proof. By Slater’s condition, there is ¯x∈C such thatG j(x)>0 for j =1, ...,r. By continuityof G j( j = 1, ...,r), there isx∗ >> x, such thatG j(x∗) > 0 for j = 1, ...,r. Thus,x∗ >> 0 andx∗ ∈C. So, the result follows directly from Corollary 8.

Remark 1. To apply Theorem 50, Corollary 8 or Corollary 9, one has to checkthatf ,G j( j = 1, ...,r) are continuously differentiable on X, and quasi-concave onRn

+. How-ever, in view of Lemma 3, it is sufficient to check that f,G j( j = 1, ...,r) are continuouslydifferentiable on X, and quasi-concave onRn

++.

11.4 The Necessity Theorem of Arrow-Enthoven

Theorem 51. Suppose f,G j( j = 1, ...,r) are continuously differentiable functions on X.Suppose Gj( j = 1, ...,r) are quasi-concave onRn

+ and there is x∗ ∈C such that Gj(x∗) > 0for j = 1, ...,r. If x∈Rn

+ solves problem(P), and for each j= 1, ...,r,∇G j(x) ≠ 0, then thereis ν ∈Rr+, such that(x, ν) satisfies the Kuhn-Tucker conditions(KT II).Proof. Define m= r +n, and for j ∈ {1, . . . ,m}, defineg j ∶ X → R by g j(x) ≡ G j(x) forj = 1, ...,r,g j(x) ≡ x j−r for j = r +1, ...,r +n.

Let E ≡E(x) be the set of indices, denoted byk, for whichgk(x) = 0. If E is the emptyset, we can clearly apply the Arrow-Hurwicz-Uzawa theorem (from Chapter 10). IfE isnon-empty, we proceed as follows.

By continuity ofG j , there is ¯x>> 0, such thatG j(x) > 0 for j = 1, ...,r. That is, there isx ∈X such thatg j(x) > 0 for all j ∈ {1, . . . ,m}.

Note that ifx ∈C, then fork ∈ E, gk(x)−gk(x) = gk(x) ≥ 0. So, by quasi-concavity ofgk, we have: (x− x)∇gk(x) ≥ 0 (11.23)

by Lemma 4.We claim, now, that for eachk ∈ E, (x− x)∇gk(x) > 0. To see this, given anyk ∈

E, definexk(θ) = [x− θ∇gk(x)] whereθ > 0 and sufficiently small to makexk(θ) ≥ 0,


andG j(xk(θ)) ≥ 0 for all j ∈ {1, ...,r}. [Since x >> 0 andG j(x) > 0 for j = 1, ...,r, this ispossible]. Thusxk(θ) ∈C, and so by using (11.23), we have:

[xk(θ)− x]∇gk(x) ≥ 0 (11.24)

But (11.24) implies that:

[x− x]∇gk(x)−θ∥∇gk(x)∥2 ≥ 0 (11.25)

and so(x− x)∇gk(x) > 0, since∇gk(x) ≠ 0.Defineh= (x− x). Thenh∇gk(x) > 0 for all k ∈E, and we can again apply the Arrow-

Hurwicz-Uzawa theorem (from Chapter 10). Thus, we have shownthat whetherE isempty or not, we can apply the Arrow-Hurwicz-Uzawa theorem to getλ = (ν, µ) ∈Rr+×Rn

+,such that

(i) ∇ f (x)+ r∑j=1

ν j∇G j(x)+ µ= 0

(ii) νG(x)+ µx= 0.From (i), andµ≥ 0, we get

∇ f (x)+ r∑j=1

ν j∇G j(x) ≤ 0

From (ii) and[x,G(x)] ≥ 0,[µ, ν] ≥ 0, we get(iii) µx= 0(iv) νG(x) = 0

Multiplying (i) by x and using (iii), we get

(v) x[∇ f (x)+ r∑j=1

ν j∇G j(x)] = 0

Thus, the Kuhn-Tucker conditions(KT II) are satisfied by(x, ν) ∈Rn+×Rr+.



Problem 45(Applying Arrow-Enthoven’s theory of Quasi-Concave Programming).

Let a,b be arbitrary positive real numbers. Consider the following constrained opti-mization problem:

Maximize x1x2+ [x1/(1+x1)]sub ject to ax1+bx2 ≤ 1(x1,x2) ∈R2+

⎫⎪⎪⎪⎬⎪⎪⎪⎭(P)(a) DefineX = {(x1,x2) ∈R2 ∶ x1 > −1,x2 > −1}. Let f ∶X→R andg ∶X→R be defined by:f (x1,x2) = x1x2+ [x1/(1+x1)], g(x1,x2) = (1−ax1−bx2) for all (x1,x2) ∈X. Verify thatXis an open set inR2, and thatf andg are continuously differentiable onX. Show thatfandg are quasi-concave onR2+ , and thatf is not concave onR2+ .

(b) Write down the Kuhn-Tucker conditions for problem(P) along the lines of Arrow-Enthoven for a pair(x, λ) ∈R2+×R+.

(c) Show that if(x, λ) ∈R2+×R+ satisfies the Kuhn-Tucker conditions in (b) above, then(i) (1−ax1−bx2) = 0 ; (ii) x1 > 0 ; (iii) ∇ f (x) ≠ 0 , and (iv)λ > 0.

(d) Define a functionh ∶R+→R as follows:

h(z) = [b/(1+z)2]−2az+1 f or z≥ 0

Show that there is a unique positive solution to the equation: h(z) = 0. Call this solutionc.(e) Use the Arrow-Enthoven sufficiency theorem to show that (i) if c > (1/a), then

x= ((1/a),0) solves problem(P); (ii) if c≤ (1/a), thenx= (c,(1−ac)/b) solves problem(P).Solution.

(a) Define the setX = {(x1,x2) ∈R2 ∣ x1 > −1 andx2 > −1}

For any(x1, x2) ∈X, let r =min{x1+1, x2+1}. Then we haveB(x,r) ⊂X, soX is anopen set inR2. Also, we have thatX ⊃R2+.

Define f ∶X→R andg∶X→R by

f (x1,x2) = x1x2+ x1

1+x1for all (x1,x2) ∈X

g(x1,x2) = 1−ax1−bx2 for all (x1,x2) ∈X


The functionsf andg are continuously differentiable onX.

Note that we only have to establish quasi-concavity onR2+. Sinceg is linear onR2+,it is quasi-concave onR2+. For allx ∈R2++, the bordered Hessian off atx is

Bf (x) =⎡⎢⎢⎢⎢⎢⎢⎣

0 x2+ 1(1+x1)2 x1

x2+ 1(1+x1)2

−2(1+x1)3 1

x1 1 0

⎤⎥⎥⎥⎥⎥⎥⎦The second leading principal minor ofBf (x) is −(x2+ 1

(1+x1)2)2 < 0 for all x ∈R2++.

And for all x ∈R2++, the determinant ofBf (x) is

detBf (x) = −(x2+ 1(1+x1)2)(−x1)+x1(x2+ 1(1+x1)2 + 2x1(1+x1)3)= 2x1x2+ 2x1(1+x1)2 +

2x21(1+x1)3

> 0

Therefore f is quasi-concave onR2++. Since f is continuous onR2+ and quasi-concave onR2++, by Lemma 3, Chapter 11,f is quasi-concave onR2+. But forall x∈R2++, the determinant of the Hessian off atx is−1≱ 0, so the Hessian off atxis not negative semi-definite, which means thatf is not concave onR2++. Thereforef is not concave onR2+.

Since f is quasi-concave but not concave onX, we need to verify two further con-ditions in order to apply the Arrow-Enthoven Sufficiency Theorem. In part (b) wewill show that if a pair(x, λ) ∈ R2+ ×R+ satisfies the Kuhn-Tucker conditions, then∇ f (x) ≠ 0. Slater’s Conditions holds because(0,0) ∈R2+ andg(0,0) = 1> 0.

Therefore all the conditions of the Arrow-Enthoven Sufficiency Theorem are met,so if some pair(x, λ) ∈R2+×R+ satisfies the Kuhn-Tucker conditions, then ˆx solvesproblem(P).

(b) A pair (x, λ) ∈ R2+ ×R+ satisfies the Kuhn-Tucker conditions for problem(P) if it


satisfies the following:

x2+ 1(1+ x1)2 −aλ ≤ 0 (5.1)

x1−bλ ≤ 0 (5.2)

x1(x2+ 1(1+ x1)2 −aλ) = 0 (5.3)

x2(x1−bλ) = 0 (5.4)

1−ax1−bx2 ≥ 0 (5.5)

λ(1−ax1−bx2) = 0 (5.6)

(c) Seeking contradiction, suppose thatλ = 0. Then by (5.2), we have ˆx1 = 0, so (5.1)givesx2 ≤ −1, which is a contradiction of ˆx2 ≥ 0. Thereforeλ > 0, so (5.6) yields

1−ax1−bx2 = 0 (5.7)

Again seeking contradiction, suppose that ˆx1 = 0. By (5.7), we have ˆx2 =1b > 0.

Therefore by (5.4) we must have ˆx1 = bλ > 0 sinceλ > 0, which is a contradiction. Soit must be that ˆx1 >0. SinceD2 f (x) = x1 andx1 >0, we have that∇ f (x) ≠0. Anotherconsequence of ˆx1 > 0 is that we can use (5.3) to write

x2+ 1(1+ x1)2 −aλ = 0 (5.8)

(d) Define the functionh∶R+→R by

h(z) = b(1+z)2 −2az+1 for all z≥ 0

Now, h is a continuous function onR+. Its derivative is

h′(z) = −2b(1+z)3 −2a< 0 for all z> 0

With continuity ofh on its domain, this shows thath is strictly decreasing onR+. Inaddition, we haveh(0) = b+1> 0 and limz→∞h(z) = −∞. By the Intermediate ValueTheorem, then, there is somec> 0 such thath(c) = 0.


(e) We can split the analysis into two cases, depending on whether (5.2) holds withequality or strict inequality.

Case 1: ˆx1−bλ < 0. Then by (5.4), we have ˆx2 = 0, so (5.7) gives ˆx1 =1a > 0. Using

this and (5.8), we can rewrite the requirement for this case as

0> x1−bλ =1a−b(1

ax2+ 1

a(1+ x2)2) = 1a− b

a(1+ 1a)2

This implies that

h(1a) = b

(1+ 1a)2 −1> 0= h(c)

Sinceh is strictly decreasing on its domain, this requiresc> 1a. Note that from (5.8),

we haveλ = a(a+1)2 > 0.

Case 2: ˆx1−bλ = 0. Then, solving (5.7) for ˆx2 and using this in (5.8), we have

1b− a

bx1+ 1(1+ x1)2 − a

bx1 = 0

Multiplying both sides byb, we have

b(1+ x1)2 −2ax1+1= 0

Sincec is the unique point inR+ that givesh(c) = 0, it must be that ˆx1 = c. Then by(5.7), we have that ˆx2 =

1−acb . Since we require ˆx2 ≥ 0, a requirement for this case is

c≤ 1a. Note thatλ = 1

bx1 =cb > 0.

Recall that we showed in parts (a) and (b) that the Arrow-Enthoven SufficiencyTheorem applies. So we have the following conclusions:

(i) If c > 1a then(x, λ) = (1

a,0,a

(a+1)2) is the unique solution to the Kuhn-Tucker

conditions, so ˆx= (1a,0) solves problem(P).

(ii) If c ≤ 1a then (x, λ) = (c, 1−ac

b ,

cb) is the unique solution to the Kuhn-Tucker

conditions, so ˆx= (c, 1−acb ) solves problem(P).


References:This material is primarily based on K.J. Arrow and A.C. Enthoven,“Quasi-concave

Programming”,Econometrica(1961), 779-800. This paper also provides a further alter-native condition (involving twice differentiability of the objective function, and a non-zerogradient vector of the objective function at ˆx) under which Theorem 1 is valid.

Theorem 2 is based on the Arrow-Hurwicz-Uzawa theorem discussed in Chapter 10.For this, see K.J. Arrow, L. Hurwicz and H. Uzawa, “ConstraintQualifications in Maxi-mization Problems”,Naval Research Logistics Quarterly, 8 (1961), 175-191.

The theory of quasi-concave programming can be further extended to the class of“pseudo-concave” objective functions [and quasi-concaveconstraint functions]. For this,seeNon-Linear Programmingby O.L. Mangasarian(Chapters 9, 10).

Chapter 12

Linear Programming

12.1 The Primal and Dual Problems

The theory of linear programming is concerned with the following problem:

(P)⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

Maxq′x q′ is1×nSub ject to Ax≤ b Aism×n

x≥ 0 xisn×1bism×1

We will call (P) thePrimal problem. Associated with(P) is the following problem:

(D)⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

Miny′b bism×1Sub ject to y′A≥ q′ Aism×n

y′ ≥ 0 y′ is1×mq′ is1×n

We will call (D) theDual problem.Define the set offeasible solutions to the Primalas

F(P) = {x≥ 0suchthat Ax≤ b}Define the set offeasible solutions to the Dualas

F(D) = {y′ ≥ 0suchthat y′A≥ q′}

193

CHAPTER 12. LINEAR PROGRAMMING 194

12.2 Optimality Criterion

In studying solutions to the Primal and Dual problems, we first establish the followingoptimality criterion: ifxo is a feasible solution to the primal andyo′ a feasible solution tothe dual, and theirvaluesare equal, [that is,q′xo = yo′b], thenxo is an optimal solution tothe Primal [that is,xo solves(P)] and yo′ is an optimal solution to the Dual [that is,yo′

solves(D)].Lemma 6. If xo ∈ F(P), yo′ ∈ F(D), then

q′xo ≤ yo′Axo ≤ yo′b (12.1)

Proof. Sinceyo′ ∈ F(D), andxo ≥ 0

q′xo ≤ yo′Axo (12.2)

Sincexo ∈ F(P), andyo′ ≥ 0yo′b≥ yo′Axo (12.3)

Combining (12.2) and (12.3) yields (12.1).

Theorem 52(Optimality Criterion). If xo ∈ F(D), and

q′xo = yo′b (12.4)

then xo solves(P) and yo′solves(D).

Proof. Let x ∈ F(P). Then sinceyo′ ∈ F(D), we can use Lemma 6 to get

q′x≤ yo′b

and we can use (12.4) to getq′x≤ q′xo. Soxo solves(P). Similarly yo′ solves(D).12.3 The Basic Duality Theorems

The basic duality theorems of Linear Programming may be stated as follows. If boththe primal and the dual have feasible solutions, then both have optimal solutions and thevalues of the optimal solutions are the same; if either the primal or the dual is infeasible,then neither has an optimal solution. We establish these results in Theorem 53 and 54below.


Lemma 7 (Non-negative solutions of linear inequalities). Exactly one of the followingalternative holds.

(1) Either the inequalityBu≤ a

has a non-negative solution for u(2) Or the inequalities

v′B≥ 0, v′a< 0

have a non-negative solution for v′.

The proof, which is omitted, follows from the Farkas Lemma; see Chapter 7 on ConvexAnalysis

Lemma 8. If there is xo in F(P), and yo′in F(D) then there isx in F(P), andy′ in F(D)

such thatq′x≥ y′b

Proof. Suppose not. ThenAx ≤ b−A′y ≤ −q−q′x+b′y ≤ 0

⎫⎪⎪⎪⎬⎪⎪⎪⎭has no non-negative solution for(x,y). That is,

⎡⎢⎢⎢⎢⎢⎣A 00 −A′−q′ b′

⎤⎥⎥⎥⎥⎥⎦[ x

y] ≤

⎡⎢⎢⎢⎢⎢⎣b−q0

⎤⎥⎥⎥⎥⎥⎦has no non-negative solution for(x,y). By Lemma 7, there is an m-vectorz′ ≥ 0, ann-vectorw′ ≥ 0, and a scalarθ ≥ 0, such that

[z′,w′, θ]⎡⎢⎢⎢⎢⎢⎣

A 00 −A′−q′ b′

⎤⎥⎥⎥⎥⎥⎦≥ 0

and

[z′w′, θ]⎡⎢⎢⎢⎢⎢⎣

b−q0

⎤⎥⎥⎥⎥⎥⎦< 0

That is,z′A≥ θq′ (12.5)


w′A′ ≤ θb′ (12.6)

z′b<w′q (12.7)

Now, we claim thatθ ≠ 0. For if θ = 0, then

z′b≥ z′Axo´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ ≥ θq′x0´¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¶ = 0

(b≥Axo) Using (12.5)

and

w′q≤w′A′yo´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ ≤ θb′y0´¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¶ = 0

(q≤A′yo) Using (12.6)

Thus,w′q ≤ z′b, which contradicts (12.7). This establishes our claim thatθ ≠ 0. Sinceθ ≥ 0, we haveθ > 0. Using (12.5),

z′

θA≥ q′, so(z′

θ) ∈ F(D).

Using (12.6),w′

θA′ ≤ b′, so(w′

θ) ∈ F(P).

Soq′ (wθ ) ≤ (z′

θ )b by Lemma 6, which contradicts (12.7) again.

Theorem 53(Fundamental Theorem of Linear Programming). If there is xo in F(P) andyo′ in F(D), then there isx in F(P) andy′ in F(D) such that

(i) q′x= y′b(ii) x solves(P), y′ solves(D).

Proof. (i) By Lemma 8, there is ˜x in F(P) andy′ in F(D), such that

q′x≥ y′b

By Lemma 6,q′x≤ y′b

Soq′x= y′b

(ii) By (i) and Theorem 52, ˜x solves(P) andy′ solves(D).


Theorem 54. If either F(P) = φ, or F(D) = φ, then(P) has no solution and(D) has nosolution.

Proof. SupposeF(P) = φ. Then(P) clearly has no solution. To show that(D) has nosolution, suppose, on the contrary, thatyo′ solves(D).

SinceAx≤ b has no non-negative solution there isy′ ≥ 0 such that

y′A≧ 0, y′b< 0

by Lemma 7. Then[yo′+y′] ∈ F(D) since

yo′+y′ ≥ 0

[yo′+y′]A≥ q′

Also, [yo′+y′]b= yo′b+y′b< yo′b

Soyo′ does not solve(D), a contradiction. A similar proof can be worked out by supposingF(D) = φ.

12.4 Complementary Slackness

We finally state a result which has been very useful in applications of linear programming.It says that ifxo andyo′ are optimal solutions to the primal and dual problems, then (i)whenever a constraint (say thejth) of the primal problem is not “binding” (atxo), the opti-mal solution of the corresponding dual variable (that is,yo

j ) must be zero; (ii) whenever aconstraint (say theith) of the dual problem is not “binding” (atyo′), the optimal solution ofthe corresponding primal variable (that isxo

i ) must be zero. We establish this in Theorem55 below.

Corollary 10. If xo solves(P) and yo′ solves(D), then q′xo = yo′Axo = yo′b.

Proof. Using Lemma 2,yo′b≥ yo′Axo ≥ q′xo. Sincexo solves(P) andyo′ solves(D), wehaveyo′b= q′xo, by Theorem 53. Henceyo′b= yo′Axo = q′xo.

Corollary 11. If xo solves(P) and yo′ solves(D), then(a) yo

j [Axo−b] j = 0 f oreach j(b) [yo′A−q′]i xo

i = 0 f oreachi


Proof. By Corollary 10,yo′[Axo−b] = 0 (12.8)

Sinceyo′ ≥ 0,Axo−b≤ 0, we obtainyo′j [Axo−b] j ≤ 0 for eachj. This, together with (12.8)

impliesyo′j [Axo−b] j = 0 for j = 1, ...,m. This proves (a). The proof of (b) is similar.

Theorem 55(Goldman-Tucker). If xo ∈ F(P), yo′ ∈ F(D), then the following two state-ments are equivalent:

(i) x0solves(P)and yo′solves(D)

(ii) (a) yoj = 0whenever[Axo] j < b j , and

(b) x0i = 0whenever[yo′A]i > qi

Proof. [(ii) implies (i)] . If [Axo] j = b j , then yoj [Axo] j = yo

j b j . If [Axo] j < b j , thenyo

j [Axo] j =0= yoj b j , by (ii) (a). Thusyo′Axo = yo′b. Similarlyyo′Axo =q′xo. Soyo′b=q′xo.

Hencexo solves(P), yo′ solves(D), by Theorem 52.[(i) implies (ii)]. Sincexo solves(P) andyo′ solves(D), we have by Corollary 11,

yoj [Axo−b] j = 0 f oreach j

and [yo′A−q′]ixoi = 0 f oreachi

Thus, whenever[Axo−b] j < 0, we must haveyoj = 0. And, whenever[yo′A−q′]i > 0, we

must havexoi = 0.



Problem 46. The theory of linear programming is concerned with the following problem:

(P)

⎧⎪⎪⎪⎨⎪⎪⎪⎩Maximize q′xsubject to Ax≤ b

and x≥ 0,

where q and x are is1×n, A is m×n and b is m×1. We will call (P) the Primal problem.Associated with (P) is the following problem:

(D)

⎧⎪⎪⎪⎨⎪⎪⎪⎩Minimize y′bsubject to y′A≥ q′

and y≥ 0,

where y is m×1. We will call (D) the Dual problem. Define the set of feasible solutions tothe Primal as

F(P) = {x ∈Rn+ ∶Ax≤ b} .

Define the set of feasible solutions to the Dual as

F(D) = {y ∈Rm+ ∶ y′A≥ q′} .

(a) Supposex ∈ F(P), y ∈ F(D) and(x, y) satisfies

y′(b−Ax) = (y′A−q′)x= 0.

Show that(x, y) satisfies the Kuhn-Tucker conditions (KT II) for problem (P).Use theArrow-Enthoven sufficiency theorem to conclude thatx solves problem (P). Show thatysolves problem (D) by using an analogous argument.(b) Supposex solves (P) andy solves (D). Using the fact thatx ∈F(P) andy∈F(D), showthat

q′x≤ y′Ax≤ y′b. (1)

Using the fact thatx solves (P) and applying the Arrow-Hurwicz-Uzawa necessity theorem,show that there existsy ∈ F(D) such that

y′(b−Ax) = (y′A−q′)x= 0. (2)

Using the fact thaty solves (D), and (2), show that q′x≥ y′b. Using this inequality and (1),show that

y′(b−Ax) = (y′A−q′)x= 0. (3)


Solution (a) Let X = Rn be an open set containingRn+. Let us definef (x) = q′x and

G j(x) = (b−Ax) j on X, j = 1, ...,m. A pair (x, y) ∈Rn+×Rm

+ satisfies the KT II conditionsfor the problem (P) if and only if

(i)(a) Di f (x)+∑mj=1 y jDiG j(x) ≤ 0 for i = 1, ...,n,(i)(b) xi(Di f (x)+∑m

j=1 y jDiG j(x)) ≤ 0 for i = 1, ...,n,(ii)(a) G j(x) ≥ 0 for j = 1, ...,m,(ii)(b) ∑mj=1 y jG j(x) = 0.

For the linear problem (P) these conditions look like

(i)(a) qi +∑mj=1 y j(−a ji) ≤ 0 for i = 1, ...,n,(i)(b) xi(qi +∑m

j=1 y j(−a ji)) = 0 for i = 1, ...,n,(ii)(a) (b−Ax) j ≥ 0 for j = 1, ...,m,(ii)(b) ∑mj=1 y j(b−Ax) j = 0.

Conditions (i)(a) can be rewritten as(q′ − y′A)i ≤ 0 and hence are satisfied because ¯y ∈F(D). Conditions (ii)(a) follows from ¯x ∈ F(P). Conditions (i)(b) are implied by (i)(a),x ∈Rn

+ and(y′A−q′)x= 0. Condition (ii)(b) follows from (ii)(a), ¯y ∈Rm+ andy′(b−Ax) = 0.

Since f and g j, j = 1, ...,m, are linear on the open setX = Rn containingRn

+, thesefunctions are continuously differentiable onRn and quasi-concave onRn

+. Since f isconcave onX, it follows that condition (c) of the Arrow-Enthoven sufficiency theorem(Theorem 50, p. 79 in the Lecture Notes) is satisfied. Therefore if a pair(x, y) ∈Rn

+×Rm+

satisfies the KT II conditions for the problem (P), then the Arrow-Enthoven sufficiencytheorem implies that ¯x solves problem (P). By the analogous argument it follows thatysolves problem (D).

(b) If x ∈ F(P) and y ∈ F(D), we haveAx ≤ b and y′A ≥ q′. Since ¯x ≥ 0 and y ≥ 0,premultiplying byy′ and postmultiplying by ¯x yield y′Ax≤ y′b andy′Ax≥ q′x.

Let us defineg j(x) = (b−Ax) j and gm+i(x) = xi on X = Rn, j = 1, ...,m, i = 1, ...,n.ThenX =Rn is open and convex set, functionsf andgk, k = 1, ...,n+m, are continuouslydifferentiable onX, and functiongk is convex onX for k= 1, ...,n+m. Hence condition (b)of the AHU constraint qualification is satisfied. Therefore,if x solves (P), then ¯x is a pointof constrained local maximum forf , and by the Arrow-Hurwicz-Uzawa necessity theorem(Theorem 48, p.73 in the Lecture Notes) there existsλ ∈Rn+m

+ such that(x, y) satisfy theKuhn-Tucker (KT I) conditions:

(i) Di f (x)+∑m+nk=1 λkDigk(x) = 0, i = 1, ...,n,

(ii) gk(x) ≥ 0, k= 1, ...,n+m,

(iii) λkgk(x) = 0, k= 1, ...,n+m.


Hence(i) qi +∑m

j=1 λ j(−a ji)+ λm+i = 0, i = 1, ...,n,(ii) (b−Ax) j ≥ 0, j = 1, ...m,

(ii ′) xk ≥ 0, k=m+1, ...,m+n,(iii) λ j(b−Ax) j = 0, j = 1, ...,m,

(iii ′) λm+i xi = 0, i = 1, ...,n.

If we definey = (λ1, ..., λm)′, then (ii) implies ˆy ∈ F(D) and (iii) implies y′(b−Ax) = 0.Also it follows from (i) that

(y′A−q′)x= n∑i=1

xi(y′A−q′)i = n∑i=1

xi⎛⎝−qi + m∑

j=1λ ja ji

⎞⎠ =n∑

i=1xi λm+i = 0.

From (2) we immediately haveq′x= y′Ax= y′b. Now, since ¯y solves (D) and ˆy ∈ F(D), wehavey′b≤ y′b, thereforeq′x = y′b≥ y′b. From this inequality and from (1) it follows thatq′x= y′Ax= y′b, hence (3) is satisfied.

Problem 47. Consider the following linear programming problem:

Maximize 3x1+4x2+3x3

subject to x1+x2+3x3 ≤ 12,2x1+4x2+x3 ≤ 42

and (x1,x2,x3) ∈R3+

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭(Q)

Find a solution to problem (Q) showing your procedure clearly. [Hint: you might wantto use your solution to problem 1 of PS11, and the duality theory for linear programmingdeveloped in problem 1 of PS12].(a) Write down the Kuhn-Tucker conditions for problem (P).(b) Use the Kuhn-Tucker (sufficiency) theorem to show thaty= (2,1/2) solves problem (P)(c) Draw an appropriate diagram to illustrate your solution.

Solution (a) If we define

q=⎛⎜⎝

343

⎞⎟⎠ , A= ( 1 1 32 4 1

) , b= ( 1242) ,

then problem (Q) and its dual problem (QD) can be written as

(Q)

⎧⎪⎪⎪⎨⎪⎪⎪⎩Maximize q′xsubject to Ax≤ b

and x ∈R3+,

(QD)

⎧⎪⎪⎪⎨⎪⎪⎪⎩Minimize y′bsubject to y′A≥ q′

and y ∈R2+.,


where problem (QD) reads as

(QD)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

Minimize 12y1+42y2

subject to y1+2y2 ≥ 3,y1+4y2 ≥ 4,3y1+y2 ≥ 3

and y ∈R2+.

Let us recall that ˆy= (2,1/2) solves problem (QD′) from problem 1 of PS11 given by

(QD′)

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

Minimize 12y1+42y2

subject to y1+2y2 ≥ 3,y1+4y2 ≥ 4

and y ∈R2+.

Sincey= (2,1/2)′ solves (QD′) and the third constraint 3y1+y2 ≥ 3 from problem (QD) issatisfied at ˆy, we can conclude that ˆy= (2,1/2) also solves the dual problem (QD). Assumex solves the primal problem (Q). Then complementary slackness conditions (Corollary 11,p. 87 in the Lecture Notes) are satisfied:

(a) y j (Ax−b) j = 0 for j = 1,2,(b) (y′A−q′)i xi = 0 for i = 1,2,3.

Since

y′A−q′ = ( 2 1/2 )( 1 1 32 4 1

)−( 3 4 3 ) = ( 0 0 7/2 ) ,and hence(y′A−q′)3 = 7/2> 0, it follows from (b) that ˆx3 = 0. Since ˆy1 > 0 andy2 > 0, wecan conclude from (a) that(Ax−b)1 = (Ax−b)2 = 0, i.e.

{ x1+ x2 = 12,2x1+4x2 = 42.

It immediately follows that ˆx1 = 3 andx2 = 9. Therefore ˆx= (3,9,0)′ is the only candidatefor the solution to problem (Q). Let us prove now that ˆx actually solves (Q) by checkingcondition (ii) of the Goldman-Tucker theorem (Theorem 55, p. 87 in the Lecture Notes).Since

Ax−b= 0 and y′A−q′ = (0,0,7/2) ,it follows that (ii ,a) y j = 0 whenever(Ax) j < b j (never),(ii ,b) xi = 0 whenever(y′A)i > qi (for i = 3),


Therefore Goldman-Tucker theorem implies that ˆx = (3,9,0)′ solves problem (Q). [Pleasenote that since in this problem we haveai j > 0 for all i and j, it immediately followsthat constraint set for problem (Q) is bounded, and hence theexistence of the solution toproblem (Q) follows form the Weierstrass theorem. In general, some elements ofA maybe non-positive, and applicability of the Weierstrass theorem might be more difficult tojustify.]


References:This material is based almost entirely on D. Gale,The Theory of LinearEconomic Models, McGraw-Hill, New York, 1960, chapter 3.

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